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What is the truth value of f(x)=y if x is outside the domain of the function f?
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Dan Christensen
Dec 6, 2021, 12:54:54 AM12/6/21
to
Believe it or not, this is actually a point of controversy to some here. A related question might be, given an indexed array of 10 Boolean (true-or-false) variables, what is the truth value of the 11th one? It seems to me that in both cases, the truth value is undefined.
Your comments?
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Your comments?
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Mostowski Collapse
Dec 6, 2021, 2:31:01 AM12/6/21
to
Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
Now take contraposition you get:
ALL(a):ALL(b):[~P(a) | ~Q(b) => ~F(a,b)]
Or in words, if (a,b) is outside of the domain and co-domain, then F is false.
https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
Now take contraposition you get:
ALL(a):ALL(b):[~P(a) | ~Q(b) => ~F(a,b)]
Or in words, if (a,b) is outside of the domain and co-domain, then F is false.
https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
Serg io
Dec 6, 2021, 10:21:53 AM12/6/21
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look at inverse sine, cos, sec, cosec, functions etc...
also there are restrictions on range...
Serg io
Dec 6, 2021, 10:26:56 AM12/6/21
to
Fritz Feldhase
Dec 6, 2021, 11:48:49 AM12/6/21
to
On Monday, December 6, 2021 at 6:54:54 AM UTC+1, Dan Christensen wrote:
> Your comments?
JB seems to be obsessed by your DC Proof. Many of his "arguments" are just "higher nonsense". Completely pointless.
Though you are an ignorant idiot too.
> Your comments?
JB seems to be obsessed by your DC Proof. Many of his "arguments" are just "higher nonsense". Completely pointless.
Though you are an ignorant idiot too.
Mostowski Collapse
Dec 6, 2021, 12:42:38 PM12/6/21
to
Fritz Feldhase stick it up your ass and go fuck yourself.
Can you point some "higher nonsense" that I posted?
For example this here is no "higher nonsense":
ALL(x):[A(x) => B(x)], its actually the definition of ⊆.
And this here is no "higher nonsense":
yourself, A(x) => B(x) is logically equivalent to ~B(x) => ~A(x).
https://en.wikipedia.org/wiki/Contraposition#Intuitive_explanation
> Or in words, if (a,b) is outside of the domain and co-domain, then F is false.
The nonsense is rather on your side Fritz Feldhase. I suspect that you
are the former Crank "Me", right? Otherwise I do not understand this
attack, than a retaliation, since Crank "Me" had also problems with functions.
LMAO!
Can you point some "higher nonsense" that I posted?
For example this here is no "higher nonsense":
> Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
> ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
Explanation: Just the usual translation of A ⊆ B into
> ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
ALL(x):[A(x) => B(x)], its actually the definition of ⊆.
And this here is no "higher nonsense":
> Now take contraposition you get:
> ALL(a):ALL(b):[~P(a) | ~Q(b) => ~F(a,b)]
Explanation Unless you deny contraposition. But you can verify contraposition
> ALL(a):ALL(b):[~P(a) | ~Q(b) => ~F(a,b)]
yourself, A(x) => B(x) is logically equivalent to ~B(x) => ~A(x).
https://en.wikipedia.org/wiki/Contraposition#Intuitive_explanation
> Or in words, if (a,b) is outside of the domain and co-domain, then F is false.
are the former Crank "Me", right? Otherwise I do not understand this
attack, than a retaliation, since Crank "Me" had also problems with functions.
LMAO!
Mostowski Collapse
Dec 6, 2021, 12:45:53 PM12/6/21
to
Now we have a new Crank "Me", Fritz Feldhase, one more
troubled soul, that doesn't understand functions.
Jim Burns
Dec 6, 2021, 2:21:19 PM12/6/21
to
On 12/6/2021 11:48 AM, Fritz Feldhase wrote:
> On Monday, December 6, 2021 at 6:54:54 AM UTC+1,
> Dan Christensen wrote:
>> Your comments?
>
> JB seems to be obsessed by your DC Proof.
JB == Jim Burns?
> On Monday, December 6, 2021 at 6:54:54 AM UTC+1,
> Dan Christensen wrote:
>> Your comments?
>
> JB seems to be obsessed by your DC Proof.
What leads you to think I'm obsessed with DC Proof?
"Obsessed with WM" seems much more apt.
I don't think I am, but if I am, there are much worse
things to be obsessed with. WM springs to mind.
> Many of his "arguments" are just "higher nonsense".
Do you think he should care?
I wonder if "higher nonsense" is intended as an insult.
It's been a few years, but I think one of my math
professors referred to the axiom of choice as
"lovely nonsense". And, if we ever needed to call upon
its awesome power (not so often as you might think),
its use was pointed out, and the theorems that followed
had a not-completely-legitimate aroma about them.
But that was in an earlier, more innocent age, the 1970s.
I don't want to flatter myself, but...
|
| You can recognize a small truth because its opposite
| is a falsehood. The opposite of a great truth is
| another truth.
|
-- Niels Bohr.
> Completely pointless.
You and David Petry should get together for a nice chat.
He also thinks mathematics should have a _purpose_
> Though you are an ignorant idiot too.
Archimedes Plutonium
Dec 6, 2021, 2:34:32 PM12/6/21
to
Dan, a function, true function is defined in True Math as never being outside the domain.
This is one of the reasons the only true functions are polynomial functions.
This is one of the reasons the only true functions are polynomial functions.
Mostowski Collapse
Dec 6, 2021, 2:50:07 PM12/6/21
to
Fritz Feldhase is possibly Crank "Me", which knows me
by my name Jan Burse. Dan-O-Matik is also fond of always
repeating my name which you can see in my email.
But just refer to me as MC, since I use the nickname
Mostowski Collapse here, then there is no confusion
with JB, which would be Jim Burns, right?
To put ZFC into the spheres of higher nonsense is possibly
the work of WM and Dan-O-Matik, WM with extra steps,
himself. They always mystify ZFC, but
we are dealing here with beginner math.
by my name Jan Burse. Dan-O-Matik is also fond of always
repeating my name which you can see in my email.
But just refer to me as MC, since I use the nickname
Mostowski Collapse here, then there is no confusion
with JB, which would be Jim Burns, right?
To put ZFC into the spheres of higher nonsense is possibly
the work of WM and Dan-O-Matik, WM with extra steps,
himself. They always mystify ZFC, but
we are dealing here with beginner math.
Alan Mackenzie
Dec 6, 2021, 4:35:00 PM12/6/21
to
Mostowski Collapse <burs...@gmail.com> wrote:
> Its really beginner math.
> An easy corrolar of doing it properly is that if G is finite,
> then * and -1 are both also finite. Dan-O-Matik claimed to know what is
> finite somewhere else, by invoking Dedekind.
Why have you been expressing such disparagement of Dan in the last few
weeks?
> Can Dan-O-Matik absolutely prove, that if the object G is finite,
> and <G,*,-1> is a group ....
Er, doesn't a group have just two (not three) components; the set and
the operation?
> .... that then * is finite and -1 is finite? I doubt that Dan-O-Matik
> can prove that with his nonsense.
How can *, an operator, be finite? That seems like a category error,
like saying a mathematical operator is orange. -1 is finite, for any
reasonable definitions of -1 and finite.
> In his nonsense world * can belong to the group G ....
The operator *, as you tried to specify above, does indeed "belong" to
the group <G, *>.
> .... or to some super group G', if the group G is finite the super
> group G' might be nevertheless infinite,
Indeed.
> so that * is both finite and infinite, or what?
Category error, again. Sets are finite or infinite, not operators.
> LoL Yikes, and here we have it again, dark booleans, and now dark
> groups. WM with extra steps.
And now we have empty non-sensical personal disparagement. Why? What
have you got against Dan?
--
Alan Mackenzie (Nuremberg, Germany).
> Its really beginner math.
> An easy corrolar of doing it properly is that if G is finite,
> then * and -1 are both also finite. Dan-O-Matik claimed to know what is
> finite somewhere else, by invoking Dedekind.
Why have you been expressing such disparagement of Dan in the last few
weeks?
> Can Dan-O-Matik absolutely prove, that if the object G is finite,
> and <G,*,-1> is a group ....
Er, doesn't a group have just two (not three) components; the set and
the operation?
> .... that then * is finite and -1 is finite? I doubt that Dan-O-Matik
> can prove that with his nonsense.
How can *, an operator, be finite? That seems like a category error,
like saying a mathematical operator is orange. -1 is finite, for any
reasonable definitions of -1 and finite.
> In his nonsense world * can belong to the group G ....
The operator *, as you tried to specify above, does indeed "belong" to
the group <G, *>.
> .... or to some super group G', if the group G is finite the super
> group G' might be nevertheless infinite,
Indeed.
> so that * is both finite and infinite, or what?
Category error, again. Sets are finite or infinite, not operators.
> LoL Yikes, and here we have it again, dark booleans, and now dark
> groups. WM with extra steps.
And now we have empty non-sensical personal disparagement. Why? What
have you got against Dan?
--
Alan Mackenzie (Nuremberg, Germany).
Mostowski Collapse
Dec 6, 2021, 7:06:54 PM12/6/21
to
Dan-O-Matik should try proving with his nonsense axioms.
He will possibly be able to prove something. But he will
not be able to make a definition:
Finite(S) <=> ....
Where somebody can plug-in the operator *, and then
ask Finite(*). His solution will be something where
he has to carry around further attributes, instead
of using the proper F : X -> Y function spaces,
as defined on wikipedia. Not using the proper
function spaces as on wikipedia is non-standard
and leads to clumsy proofs. Its not what is found
in today text books when we talk about a mathematical
structure such as a group.
Mostowski Collapse schrieb:
> An operator * : G x G -> G is finite, if its extension is finite.
>
> For example this operator * : {0,1} x {0,1} -> {0,1}:
>
> * | 0 1
> ------------
> 0 | 0 1
> 1 | 1 1
>
> Is finite, it is only 4 pairs. On the other hand * : R x R -> R is infinite.
He will possibly be able to prove something. But he will
not be able to make a definition:
Finite(S) <=> ....
Where somebody can plug-in the operator *, and then
ask Finite(*). His solution will be something where
he has to carry around further attributes, instead
of using the proper F : X -> Y function spaces,
as defined on wikipedia. Not using the proper
function spaces as on wikipedia is non-standard
and leads to clumsy proofs. Its not what is found
in today text books when we talk about a mathematical
structure such as a group.
Mostowski Collapse schrieb:
> An operator * : G x G -> G is finite, if its extension is finite.
>
> For example this operator * : {0,1} x {0,1} -> {0,1}:
>
> * | 0 1
> ------------
> 0 | 0 1
> 1 | 1 1
>
> Is finite, it is only 4 pairs. On the other hand * : R x R -> R is infinite.
Archimedes Plutonium
Dec 6, 2021, 7:53:32 PM12/6/21
to
The domain and range of a function are the same as the definition of function.
Your question is as foolish as asking can you have the heart and lungs outside the body and the liver 2 km away from the body and still be a human.
Some fools think that trigonometry can have the x axis be angles and the y axis be Reals or fruitcake
No, the function and domain are part and parcel the same thing.
This is another reason, the only true functions are polynomials, because the moment you state a polynomial function, it must cover all the Decimal Grid Numbers on the x-axis. And the y-axis must be the same.
So a question such as Dan's belies the fact that Dan does not understand what a function is in the first place.
Your question is as foolish as asking can you have the heart and lungs outside the body and the liver 2 km away from the body and still be a human.
Some fools think that trigonometry can have the x axis be angles and the y axis be Reals or fruitcake
No, the function and domain are part and parcel the same thing.
This is another reason, the only true functions are polynomials, because the moment you state a polynomial function, it must cover all the Decimal Grid Numbers on the x-axis. And the y-axis must be the same.
So a question such as Dan's belies the fact that Dan does not understand what a function is in the first place.
Dan Christensen
Dec 6, 2021, 7:55:56 PM12/6/21
to
On Monday, December 6, 2021 at 7:06:54 PM UTC-5, Mostowski Collapse wrote:
> Dan-O-Matik should try proving with his nonsense axioms.
> He will possibly be able to prove something. But he will
> not be able to make a definition:
>
> Finite(S) <=> ....
>
Liar. Have done so many times. See https://dcproof.wordpress.com/2014/09/17/infinity-the-story-so-far/
> Dan-O-Matik should try proving with his nonsense axioms.
> He will possibly be able to prove something. But he will
> not be able to make a definition:
>
> Finite(S) <=> ....
>
Mostowski Collapse
Dec 6, 2021, 8:00:25 PM12/6/21
to
An operator * : G x G -> G is finite, if its extension is finite.
For example this operator * : {0,1} x {0,1} -> {0,1}:
* | 0 1
------------
0 | 0 1
1 | 1 1
Is finite, it is only 4 pairs. On the other hand * : R x R -> R is infinite.
For example this operator * : {0,1} x {0,1} -> {0,1}:
* | 0 1
------------
0 | 0 1
1 | 1 1
Is finite, it is only 4 pairs. On the other hand * : R x R -> R is infinite.
Jim Burns
Dec 6, 2021, 8:01:48 PM12/6/21
to
On 12/6/2021 2:49 PM, Mostowski Collapse wrote:
> Fritz Feldhase is possibly Crank "Me", which knows me
> by my name Jan Burse. Dan-O-Matik is also fond of always
> repeating my name which you can see in my email.
Oops. I knew that.
> Fritz Feldhase is possibly Crank "Me", which knows me
> by my name Jan Burse. Dan-O-Matik is also fond of always
> repeating my name which you can see in my email.
Sorry.
Fritz Feldhase
Dec 6, 2021, 8:02:18 PM12/6/21
to
On Monday, December 6, 2021 at 8:21:19 PM UTC+1, Jim Burns wrote:
> On 12/6/2021 11:48 AM, Fritz Feldhase wrote:
> > On Monday, December 6, 2021 at 6:54:54 AM UTC+1,
> > Dan Christensen wrote:
> > >
> > > Your comments?
> > >
> > JB seems to be obsessed by your DC Proof.
> >
> JB == Jim Burns?
My bad. Sorry! JB == Jan Burse.
> On 12/6/2021 11:48 AM, Fritz Feldhase wrote:
> > On Monday, December 6, 2021 at 6:54:54 AM UTC+1,
> > Dan Christensen wrote:
> > >
> > > Your comments?
> > >
> > JB seems to be obsessed by your DC Proof.
> >
> JB == Jim Burns?
> What leads you to think I'm obsessed with DC Proof?
> "Obsessed with WM" seems much more apt.
> Oh, good. Back on topic.
Sincere apologies once again.
Dan Christensen
Dec 6, 2021, 8:04:36 PM12/6/21
to
On Monday, December 6, 2021 at 2:31:01 AM UTC-5, Mostowski Collapse wrote:
> Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
>
> ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
>
> Now take contraposition you get:
>
> ALL(a):ALL(b):[~P(a) | ~Q(b) => ~F(a,b)]
>
> Or in words, if (a,b) is outside of the domain and co-domain, then F is false.
>
This seems to be a weird artifact of the representation of a function that you have chosen. Thanks for bringing it to our attention, but AFAIK you cannot obtain this wonky result with the more standard representation: ALL(a):[a in P => F(a) in Q]
> Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
>
> ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
>
> Now take contraposition you get:
>
> ALL(a):ALL(b):[~P(a) | ~Q(b) => ~F(a,b)]
>
> Or in words, if (a,b) is outside of the domain and co-domain, then F is false.
>
In group theory, for example, we can have at a subgroup H < G such that:
(1) ALL(a): ALL(b):[a in G & b in G => a*b in G]
(2) ALL(a): ALL(b):[a in H & b in H => a*b in H]
By your reasoning: x not in H would imply that x*y=z would be false for every y and z. An obvious non-starter.
Mostowski Collapse
Dec 6, 2021, 8:17:13 PM12/6/21
to
Its beginner math!
Alan Mackenzie just read the wikipedia definition
of a function. Then it should be easy to understand
how we can determine its cardinality:
Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
And now you got it, your operator is suddently a set.
Hope this is not some news for you, because then
I have bad news for, the list of those who don't
know here what a function is (graph) is,
has just increased by one.
Corr.:
>> For example this operator * : {0,1} x {0,1} -> {0,1}:
>>
>> * | 0 1
>> ------------
>> 0 | 0 1
>> 1 | 1 1
>>
>> Is finite, it is only 4 pairs.
Its actually 4 pairs that contain pairs as their
first argument, we could also say its 4 triples:
((0,0),0)
((0,1),1)
((1,0),1)
((1,1),1)
Mostowski Collapse schrieb:
Alan Mackenzie just read the wikipedia definition
of a function. Then it should be easy to understand
how we can determine its cardinality:
Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
And now you got it, your operator is suddently a set.
Hope this is not some news for you, because then
I have bad news for, the list of those who don't
know here what a function is (graph) is,
has just increased by one.
Corr.:
>> For example this operator * : {0,1} x {0,1} -> {0,1}:
>>
>> * | 0 1
>> ------------
>> 0 | 0 1
>> 1 | 1 1
>>
>> Is finite, it is only 4 pairs.
first argument, we could also say its 4 triples:
((0,0),0)
((0,1),1)
((1,0),1)
((1,1),1)
Mostowski Collapse schrieb:
Dan Christensen
Dec 6, 2021, 8:28:49 PM12/6/21
to
On Monday, December 6, 2021 at 2:34:32 PM UTC-5, Archimedes Plutonium wrote:
> Dan, a function, true function is defined in True Math as never being outside the domain.
OK.
> Dan, a function, true function is defined in True Math as never being outside the domain.
> This is one of the reasons the only true functions are polynomial functions.
Dan
Mostowski Collapse
Dec 6, 2021, 8:29:30 PM12/6/21
to
Its really beginner math.
An easy corrolar of doing it properly is that if G is finite,
then * and -1 are both also finite. Dan-O-Matik claimed to know what is
finite somewhere else, by invoking Dedekind.
An easy corrolar of doing it properly is that if G is finite,
then * and -1 are both also finite. Dan-O-Matik claimed to know what is
finite somewhere else, by invoking Dedekind.
Can Dan-O-Matik absolutely prove, that if the object G is finite,
and <G,*,-1> is a group that then * is finite and -1 is finite?
I doubt that Dan-O-Matik can prove that with his nonsense.
In his nonsense world * can belong to the group G
or to some super group G', if the group G is finite the
super group G' might be nevertheless infinite,
so that * is both finite and infinite, or what? LoL Yikes,
super group G' might be nevertheless infinite,
and here we have it again, dark booleans, and
now dark groups. WM with extra steps.
now dark groups. WM with extra steps.
Archimedes Plutonium
Dec 6, 2021, 8:33:12 PM12/6/21
to
On Monday, December 6, 2021 at 1:34:32 PM UTC-6, Archimedes Plutonium wrote:
> Dan, a function, true function is defined in True Math as never being outside the domain.
> This is one of the reasons the only true functions are polynomial functions.
Maybe I did not make myself clear enough here. The entire point of a function in mathematics is that it must cover, it must indulge, it must activate every number on the x axis and every number on the y axis. This is not negotiable. Both axes the same numbers. Those are Decimal Grid Number systems such as 10 Grid, 100 Grid etc etc.
> Dan, a function, true function is defined in True Math as never being outside the domain.
> This is one of the reasons the only true functions are polynomial functions.
And the only numbers that exist are Decimal Grid Numbers, no negatives, only positives in 1st Grid only.
And the only functions that exist are Polynomials.
Now when you disobey these rules like Old Math with their silly obnoxious trigonometry, they went ahead and thought negative numbers were okay and had 3 more quadrants and so silly stupid were they, that they even said, -- "Hey fellow obnoxious fools, let us make the x axis be angles and let us then make the y axis be Reals of Old Math. So when you have loose marbled brain kazooo idiots doing mathematics, all hell and then some is allowed in this circus zoo known as Old Math.
When the only true numbers are Decimal Grid numbers and the only true functions are polynomials, this forbids ever the question of "Not covering a function with a domain number". Every polynomial has to answer to a x-axis number in True Mathematics.
AP, King of Science, especially Physics
Message has been deleted
Dan Christensen
Dec 6, 2021, 8:40:20 PM12/6/21
to
On Monday, December 6, 2021 at 8:33:12 PM UTC-5, Archimedes Plutonium wrote:
> On Monday, December 6, 2021 at 1:34:32 PM UTC-6, Archimedes Plutonium wrote:
> > Dan, a function, true function is defined in True Math as never being outside the domain.
> > This is one of the reasons the only true functions are polynomial functions.
> Maybe I did not make myself clear enough here. The entire point of a function in mathematics is that it must cover, it must indulge, it must activate every number on the x axis and every number on the y axis. This is not negotiable. Both axes the same numbers. Those are Decimal Grid Number systems such as 10 Grid, 100 Grid etc etc.
This is what I get for making eye contact with a crazy person! Bye....
> On Monday, December 6, 2021 at 1:34:32 PM UTC-6, Archimedes Plutonium wrote:
> > Dan, a function, true function is defined in True Math as never being outside the domain.
> > This is one of the reasons the only true functions are polynomial functions.
> Maybe I did not make myself clear enough here. The entire point of a function in mathematics is that it must cover, it must indulge, it must activate every number on the x axis and every number on the y axis. This is not negotiable. Both axes the same numbers. Those are Decimal Grid Number systems such as 10 Grid, 100 Grid etc etc.
Archimedes Plutonium
Dec 6, 2021, 10:06:25 PM12/6/21
to
Go get 'em Jan, ride him ragged, for Dan is ultra-low-insane.
Dan Christensen
Dec 7, 2021, 1:00:34 AM12/7/21
to
(Correction)
On Monday, December 6, 2021 at 2:32:37 AM UTC-5, Mostowski Collapse wrote:
> Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
>
> ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
>
> Now take contraposition you get:
>
> ALL(a):ALL(b):[~P(a) | ~Q(b) => ~F(a,b)]
>
> Or in words, if (a,b) is outside of the domain and co-domain, then F is false.
>
> https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
Rethinking the notion of a function just a bit...
Define a set of ordered pairs F such that ALL(a):[a in P => EXIST(b):[b in Q & (a,b) in F & ALL(c):[c in Q => [(a,c) in F => c=b]]]].
Note that this does NOT require that ALL(a):ALL(b):[(a, b) in F => a in P & b in Q]. The is does NOT rule out other terms in F.
Comments?
On Monday, December 6, 2021 at 2:32:37 AM UTC-5, Mostowski Collapse wrote:
> Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
>
> ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
>
> Now take contraposition you get:
>
> ALL(a):ALL(b):[~P(a) | ~Q(b) => ~F(a,b)]
>
> Or in words, if (a,b) is outside of the domain and co-domain, then F is false.
>
> https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
Define a set of ordered pairs F such that ALL(a):[a in P => EXIST(b):[b in Q & (a,b) in F & ALL(c):[c in Q => [(a,c) in F => c=b]]]].
Note that this does NOT require that ALL(a):ALL(b):[(a, b) in F => a in P & b in Q]. The is does NOT rule out other terms in F.
Comments?
Message has been deleted
Archimedes Plutonium
Dec 7, 2021, 2:04:16 AM12/7/21
to
On Tuesday, December 7, 2021 at 12:00:34 AM UTC-6, Dan Christensen wrote:
> (Correction)
> On Monday, December 6, 2021 at 2:32:37 AM UTC-5, Mostowski Collapse wrote:
> > Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
> >
> > ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
> >
> > Now take contraposition you get:
> >
> > ALL(a):ALL(b):[~P(a) | ~Q(b) => ~F(a,b)]
> >
> > Or in words, if (a,b) is outside of the domain and co-domain, then F is false.
> >
> > https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
> Rethinking the notion of a function just a bit...
>
> Define a set of ordered pairs F such that ALL(a):[a in P => EXIST(b):[b in Q & (a,b) in F & ALL(c):[c in Q => [(a,c) in F => c=b]]]].
>
> Note that this does NOT require that ALL(a):ALL(b):[(a, b) in F => a in P & b in Q]. The is does NOT rule out other terms in F.
>
> Comments?
You ask for comments, but your feeble brain can never comprehend a comment.
> (Correction)
> On Monday, December 6, 2021 at 2:32:37 AM UTC-5, Mostowski Collapse wrote:
> > Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
> >
> > ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
> >
> > Now take contraposition you get:
> >
> > ALL(a):ALL(b):[~P(a) | ~Q(b) => ~F(a,b)]
> >
> > Or in words, if (a,b) is outside of the domain and co-domain, then F is false.
> >
> > https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
> Rethinking the notion of a function just a bit...
>
> Define a set of ordered pairs F such that ALL(a):[a in P => EXIST(b):[b in Q & (a,b) in F & ALL(c):[c in Q => [(a,c) in F => c=b]]]].
>
> Note that this does NOT require that ALL(a):ALL(b):[(a, b) in F => a in P & b in Q]. The is does NOT rule out other terms in F.
>
> Comments?
You dunce, the very definition of function is tied up, inextricably with domain and range.
You dunce, you cannot play around with the domain or range and think you have a function.
It is like driving the car in Canada-- "hey, look over there, there is a human heart perched in that tree", "hey look over there-- there is a walrus liver flying in the sky"
Your problem Dan, is you want to do logic, but you have no logical brains to ever be in this business.
And let Jan Burse pound that down your dunce cap every half hour of the day.
Your question, DC, shows you are a failure of logic and should be drummed out of logic. Like the time you said that the derivative of a vertex can be assembled. Or your crusade to save Boole's OR with 2 OR 1 = 3 with AND as subtraction.
These are all symptoms, Dan, of a failure of Logic, a dunce of logic, yet you post day after day of more and more failure. Your a disgrace and a failure of logic. You are no better, in fact worse than John Gabriel, far worse than MitchR, in fact, you are in league with Pentcho Valev and his insanity of special relativity. Far worse than Wolfgang Muckenheim (spelling). Probably a little better than kibosh Parry M who cannot even proffer a dumb-arse theory and defend it for a decade like the dunce Dan Christensen. At least Dan knows there is a mistake when Parry M says 938 is 12% short of 945.
So, get out of sci.math, get out of academics. Go back to house painting...
Dan Christensen
Dec 7, 2021, 2:16:59 AM12/7/21
to
STUDENTS BEWARE: Don't be a victim of AP's fake math and science
On Tuesday, December 7, 2021 at 2:04:16 AM UTC-5, Archimedes Plutonium wrote:
> You ask for comments, but your feeble brain can never comprehend a comment.
>
> You dunce, the very definition of function is tied up, inextricably with domain and range.
>
STUDENTS BEWARE: Don't be a victim of AP's fake math and science
AP is a malicious internet troll who wants only to mislead and confuse you. He may not be all there, but his fake math and science can only be meant to promote failure in schools. One can only guess at his motives.
In AP's OWN WORDS here:
“Primes do not exist, because the set they were borne from has no division.”
--June 29, 2020
“The last and largest finite number is 10^604.”
--June 3, 2015
“0 appears to be the last and largest finite number”
--June 9, 2015
“0/0 must be equal to 1.”
-- June 9, 2015
“0 is an infinite irrational number.”
--June 28, 2015
“No negative numbers exist.”
--December 22, 2018
“Rationals are not numbers.”
--May 18, 2019
According to AP's “chess board math,” an equilateral triangle is a right-triangle.
--December 11, 2019
Which could explain...
“The value of sin(45 degrees) = 1.”
--May 31, 2019
AP deliberately and repeatedly presented the truth table for OR as the truth table for AND:
“New Logic
AND
T & T = T
T & F = T
F & T = T
F & F = F”
--November 9, 2019
AP seeks aid of Russian agents to promote failure in schools:
"Please--Asking for help from Russia-- russian robots-- to create a new, true mathematics [sic]. What I like for the robots to do, is list every day, about 4 Colleges ( of the West) math dept, and ask why that math department is teaching false and fake math, and if unable to change to the correct true math, well, simply fire that math department until they can find professors who recognize truth in math from fakery...."
--November 9, 2017
And if that wasn't weird enough...
“The totality, everything that there is [the universe], is only 1 atom of plutonium [Pu]. There is nothing outside or beyond this one atom of plutonium.”
--April 4, 1994
“The Universe itself is one gigantic big atom.”
--November 14, 2019
AP's sinister Atom God Cult of Failure???
“Since God-Pu is marching on.
Glory! Glory! Atom Plutonium!
Its truth is marching on.
It has sounded forth the trumpet that shall never call retreat;
It is sifting out the hearts of people before its judgment seat;
Oh, be swift, my soul, to answer it; be jubilant, my feet!
Our God-Pu is marching on.”
--December 15, 2018 (Note: Pu is the atomic symbol for plutonium)
On Tuesday, December 7, 2021 at 2:04:16 AM UTC-5, Archimedes Plutonium wrote:
> You ask for comments, but your feeble brain can never comprehend a comment.
>
> You dunce, the very definition of function is tied up, inextricably with domain and range.
>
AP is a malicious internet troll who wants only to mislead and confuse you. He may not be all there, but his fake math and science can only be meant to promote failure in schools. One can only guess at his motives.
In AP's OWN WORDS here:
“Primes do not exist, because the set they were borne from has no division.”
--June 29, 2020
“The last and largest finite number is 10^604.”
--June 3, 2015
“0 appears to be the last and largest finite number”
--June 9, 2015
“0/0 must be equal to 1.”
-- June 9, 2015
“0 is an infinite irrational number.”
--June 28, 2015
“No negative numbers exist.”
--December 22, 2018
“Rationals are not numbers.”
--May 18, 2019
According to AP's “chess board math,” an equilateral triangle is a right-triangle.
--December 11, 2019
Which could explain...
“The value of sin(45 degrees) = 1.”
--May 31, 2019
AP deliberately and repeatedly presented the truth table for OR as the truth table for AND:
“New Logic
AND
T & T = T
T & F = T
F & T = T
F & F = F”
--November 9, 2019
AP seeks aid of Russian agents to promote failure in schools:
"Please--Asking for help from Russia-- russian robots-- to create a new, true mathematics [sic]. What I like for the robots to do, is list every day, about 4 Colleges ( of the West) math dept, and ask why that math department is teaching false and fake math, and if unable to change to the correct true math, well, simply fire that math department until they can find professors who recognize truth in math from fakery...."
--November 9, 2017
And if that wasn't weird enough...
“The totality, everything that there is [the universe], is only 1 atom of plutonium [Pu]. There is nothing outside or beyond this one atom of plutonium.”
--April 4, 1994
“The Universe itself is one gigantic big atom.”
--November 14, 2019
AP's sinister Atom God Cult of Failure???
“Since God-Pu is marching on.
Glory! Glory! Atom Plutonium!
Its truth is marching on.
It has sounded forth the trumpet that shall never call retreat;
It is sifting out the hearts of people before its judgment seat;
Oh, be swift, my soul, to answer it; be jubilant, my feet!
Our God-Pu is marching on.”
--December 15, 2018 (Note: Pu is the atomic symbol for plutonium)
Mostowski Collapse
Dec 7, 2021, 3:09:55 AM12/7/21
to
For the 1000-th time, thats not the wiki definition.
Whats wrong with you? Cant you read what is written:
> Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
> ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
It says, a function F : P -> Q is:
- a binary relation F ⊆ P x Q
- serial
- functional
Why do you always sweep under the rug the "binary relation" part?
Whats wrong with you? Cant you read what is written:
> Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
> ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
It says, a function F : P -> Q is:
- a binary relation F ⊆ P x Q
- serial
- functional
Why do you always sweep under the rug the "binary relation" part?
Mostowski Collapse
Dec 7, 2021, 3:13:14 AM12/7/21
to
STUDENTS BEWARE Dan-O-Matik didn't take his medication,
sweeping under the rug parts of a definitions. Lets recap:
1) Wikipedia definition of a function:
a binary relation F ⊆ P x Q, serial on P,Q and functional on P,Q
2) Dan-O-Matiks psychotic definition:
serial on P,Q and functional on P,Q
Do you see any differents dear students?
sweeping under the rug parts of a definitions. Lets recap:
1) Wikipedia definition of a function:
a binary relation F ⊆ P x Q, serial on P,Q and functional on P,Q
2) Dan-O-Matiks psychotic definition:
serial on P,Q and functional on P,Q
Do you see any differents dear students?
Archimedes Plutonium
Dec 7, 2021, 3:35:13 AM12/7/21
to
On Tuesday, December 7, 2021 at 2:13:14 AM UTC-6, Mostowski Collapse wrote:
> STUDENTS BEWARE Dan-O-Matik didn't take his medication,
> sweeping under the rug parts of a definitions. Lets recap:
>
> 1) Wikipedia definition of a function:
> a binary relation F ⊆ P x Q, serial on P,Q and functional on P,Q
>
> 2) Dan-O-Matiks psychotic definition:
> serial on P,Q and functional on P,Q
>
Good on you Jan, I could not explain it any better than that. But you think the Dunce will listen-- no, he is deep into insanity
> STUDENTS BEWARE Dan-O-Matik didn't take his medication,
> sweeping under the rug parts of a definitions. Lets recap:
>
> 1) Wikipedia definition of a function:
> a binary relation F ⊆ P x Q, serial on P,Q and functional on P,Q
>
> 2) Dan-O-Matiks psychotic definition:
> serial on P,Q and functional on P,Q
>
Mostowski Collapse
Dec 7, 2021, 9:40:49 AM12/7/21
to
This barber riddle is for Dan-O-Matik, I am pretty sure he will be able
to prove that Fritz Klein is indeterminate, some dark boolean, because
the function do something (mathematical) F : M -> T, where M are
mathematicians and T are topics, can have of course, according
to Dan-O-Matik, also some values outside of mathematics.
"Es gäbe zwei Klassen von Mathematikern in Göttingen.
Zur ersten Klasse gehören die, die täten, was Felix Klein wollte, was ihnen
aber nicht gefiel. Zur zweiten Klasse gehörten diejenigen, die täten,
was ihnen gefiel, was aber Felix Klein nicht gefiel. Zu welcher
Klasse gehört Felix Klein? Als die Studenten stumm blieben,
meinte er, die Antwort wäre furchtbar einfach:
Felix Klein wäre gar kein Mathematiker"
"There are two classes of mathematicians in Göttingen. The first
class includes those who do what Felix Klein wanted but didn't
like. The second class included those who did what they liked
but didn't like Felix Klein. What class does Felix Klein belong to?
When the students remained silent, he said the answer would be
terribly simple: Felix Klein wouldn't be a mathematician at all"
Try it with Dan-O-Matik functions! LoL
to prove that Fritz Klein is indeterminate, some dark boolean, because
the function do something (mathematical) F : M -> T, where M are
mathematicians and T are topics, can have of course, according
to Dan-O-Matik, also some values outside of mathematics.
"Es gäbe zwei Klassen von Mathematikern in Göttingen.
Zur ersten Klasse gehören die, die täten, was Felix Klein wollte, was ihnen
aber nicht gefiel. Zur zweiten Klasse gehörten diejenigen, die täten,
was ihnen gefiel, was aber Felix Klein nicht gefiel. Zu welcher
Klasse gehört Felix Klein? Als die Studenten stumm blieben,
meinte er, die Antwort wäre furchtbar einfach:
Felix Klein wäre gar kein Mathematiker"
"There are two classes of mathematicians in Göttingen. The first
class includes those who do what Felix Klein wanted but didn't
like. The second class included those who did what they liked
but didn't like Felix Klein. What class does Felix Klein belong to?
When the students remained silent, he said the answer would be
terribly simple: Felix Klein wouldn't be a mathematician at all"
Try it with Dan-O-Matik functions! LoL
Mostowski Collapse
Dec 7, 2021, 9:41:50 AM12/7/21
to
Source of the riddle by way of Wolfgang Pauli it seems:
https://de.wikipedia.org/wiki/Ernst_Zermelo#Sonstiges
https://de.wikipedia.org/wiki/Ernst_Zermelo#Sonstiges
Michael Moroney
Dec 7, 2021, 10:11:50 AM12/7/21
to
💀 of Math and ☠️ of Physics Archimedes "Putin's Stooge" Plutonium
<plutonium....@gmail.com> blithered:
Plutonium, calling someone else illogical? How illogical is that?
failure of logic! Talk about projection!
> Like the time you said that the derivative of a vertex can be assembled.
> Or your crusade to save Boole's OR with 2 OR 1 = 3 with AND as subtraction.
Stupid Plutonium is so illogical he thinks ANYONE (other than possibly
himself) thinks AND is subtraction! How dumb and stoopid is that?
StupidPlutonium can't even get the AND or OR truth tables right!!
ANYONE else a failure of logic?
"I cannot believe how incredibly stupid Archie-Poo is. I mean rock-hard
stupid. Blazing hot mid-day sun on Mercury stupid. Surface of Venus
under 80 atmospheres of red hot carbon dioxide and sulfuric acid vapor
dehydrated for 300 million years rock-hard stupid. Stupid so stupid
that it goes way beyond the stupid we know into a whole different
sensorium of stupid. Archie-Poo is trans-stupid stupid. Meta-stupid.
Stupid so collapsed upon itself that it is within its own Schwarzschild
radius. Black hole stupid. Stupid gotten so dense and massive that no
intellect can escape. Singularity stupid. Archie-Poo emits more
stupid/second than our entire galaxy otherwise emits stupid/year.
Quasar stupid. Nothing else in the universe can be this stupid.
Archie-Poo is an oozingly putrescent primordial fragment from the
original Big Bang of Stupid, a pure essence of stupid so uncontaminated
by anything else as to be beyond the laws of physics that define
maximally extrapolated hypergeometric n-dimensional backgroundless
stupid as we can imagine it. Archie-Poo is Planck stupid, a quantum
foam of stupid, a vacuum decay of stupid, a grand unified theory of
stupid.
Archie-Poo is the epitome of stupidity, the epiphany of stupid, the
apotheosis of stupidity. Archie-poo is stooopid."
<plutonium....@gmail.com> blithered:
> On Tuesday, December 7, 2021 at 12:00:34 AM UTC-6, Dan Christensen wrote:
>> (Correction)
>> On Monday, December 6, 2021 at 2:32:37 AM UTC-5, Mostowski Collapse wrote:
>>> Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
>>>
>>> ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
>>>
>>> Now take contraposition you get:
>>>
>>> ALL(a):ALL(b):[~P(a) | ~Q(b) => ~F(a,b)]
>>>
>>> Or in words, if (a,b) is outside of the domain and co-domain, then F is false.
>>>
>>> https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
>> Rethinking the notion of a function just a bit...
>>
>> Define a set of ordered pairs F such that ALL(a):[a in P => EXIST(b):[b in Q & (a,b) in F & ALL(c):[c in Q => [(a,c) in F => c=b]]]].
>>
>> Note that this does NOT require that ALL(a):ALL(b):[(a, b) in F => a in P & b in Q]. The is does NOT rule out other terms in F.
>>
>> Comments?
>
> You ask for comments, but your feeble brain can never comprehend a comment.
>
> You dunce, the very definition of function is tied up, inextricably with domain and range.
>
> You dunce, you cannot play around with the domain or range and think you have a function.
>
> It is like driving the car in Canada-- "hey, look over there, there is a human heart perched in that tree", "hey look over there-- there is a walrus liver flying in the sky"
>
> Your problem Dan, is you want to do logic, but you have no logical brains to ever be in this business.
What's this? The most illogical person on the planet, Archimedes
>> (Correction)
>> On Monday, December 6, 2021 at 2:32:37 AM UTC-5, Mostowski Collapse wrote:
>>> Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
>>>
>>> ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
>>>
>>> Now take contraposition you get:
>>>
>>> ALL(a):ALL(b):[~P(a) | ~Q(b) => ~F(a,b)]
>>>
>>> Or in words, if (a,b) is outside of the domain and co-domain, then F is false.
>>>
>>> https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
>> Rethinking the notion of a function just a bit...
>>
>> Define a set of ordered pairs F such that ALL(a):[a in P => EXIST(b):[b in Q & (a,b) in F & ALL(c):[c in Q => [(a,c) in F => c=b]]]].
>>
>> Note that this does NOT require that ALL(a):ALL(b):[(a, b) in F => a in P & b in Q]. The is does NOT rule out other terms in F.
>>
>> Comments?
>
> You ask for comments, but your feeble brain can never comprehend a comment.
>
> You dunce, the very definition of function is tied up, inextricably with domain and range.
>
> You dunce, you cannot play around with the domain or range and think you have a function.
>
> It is like driving the car in Canada-- "hey, look over there, there is a human heart perched in that tree", "hey look over there-- there is a walrus liver flying in the sky"
>
> Your problem Dan, is you want to do logic, but you have no logical brains to ever be in this business.
Plutonium, calling someone else illogical? How illogical is that?
>
> And let Jan Burse pound that down your dunce cap every half hour of the day.
>
> Your question, DC, shows you are a failure of logic and should be drummed out of logic.
Again (still) amazing, failure of logic Plutonium calling others a
> And let Jan Burse pound that down your dunce cap every half hour of the day.
>
> Your question, DC, shows you are a failure of logic and should be drummed out of logic.
failure of logic! Talk about projection!
> Like the time you said that the derivative of a vertex can be assembled.
> Or your crusade to save Boole's OR with 2 OR 1 = 3 with AND as subtraction.
himself) thinks AND is subtraction! How dumb and stoopid is that?
StupidPlutonium can't even get the AND or OR truth tables right!!
>
> These are all symptoms, Dan, of a failure of Logic, a dunce of logic, yet you post day after day of more and more failure. Your a disgrace and a failure of logic.
These are all symptoms, Plutonium, of a failure of Logic, a dunce of
> These are all symptoms, Dan, of a failure of Logic, a dunce of logic, yet you post day after day of more and more failure. Your a disgrace and a failure of logic.
logic, yet you post day after day of more and more failure. Your a
disgrace and a failure of logic.
>
disgrace and a failure of logic.
>
> So, get out of sci.math, get out of academics. Go back to house painting...
Hey Ghost of Uncle Al!! What do you think of illogical Plutonium calling
ANYONE else a failure of logic?
"I cannot believe how incredibly stupid Archie-Poo is. I mean rock-hard
stupid. Blazing hot mid-day sun on Mercury stupid. Surface of Venus
under 80 atmospheres of red hot carbon dioxide and sulfuric acid vapor
dehydrated for 300 million years rock-hard stupid. Stupid so stupid
that it goes way beyond the stupid we know into a whole different
sensorium of stupid. Archie-Poo is trans-stupid stupid. Meta-stupid.
Stupid so collapsed upon itself that it is within its own Schwarzschild
radius. Black hole stupid. Stupid gotten so dense and massive that no
intellect can escape. Singularity stupid. Archie-Poo emits more
stupid/second than our entire galaxy otherwise emits stupid/year.
Quasar stupid. Nothing else in the universe can be this stupid.
Archie-Poo is an oozingly putrescent primordial fragment from the
original Big Bang of Stupid, a pure essence of stupid so uncontaminated
by anything else as to be beyond the laws of physics that define
maximally extrapolated hypergeometric n-dimensional backgroundless
stupid as we can imagine it. Archie-Poo is Planck stupid, a quantum
foam of stupid, a vacuum decay of stupid, a grand unified theory of
stupid.
Archie-Poo is the epitome of stupidity, the epiphany of stupid, the
apotheosis of stupidity. Archie-poo is stooopid."
Dan Christensen
Dec 7, 2021, 10:49:20 AM12/7/21
to
On Tuesday, December 7, 2021 at 3:09:55 AM UTC-5, Mostowski Collapse wrote:
> Dan Christensen schrieb am Dienstag, 7. Dezember 2021 um 07:00:34 UTC+1:
> > (Correction)
> > On Monday, December 6, 2021 at 2:32:37 AM UTC-5, Mostowski Collapse wrote:
> > > Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
> > >
> > > ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
> > >
> > > Now take contraposition you get:
> > >
> > > ALL(a):ALL(b):[~P(a) | ~Q(b) => ~F(a,b)]
> > >
> > > Or in words, if (a,b) is outside of the domain and co-domain, then F is false.
> > >
> > > https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
> > Rethinking the notion of a function just a bit...
> >
> > Define a set of ordered pairs F such that ALL(a):[a in P => EXIST(b):[b in Q & (a,b) in F & ALL(c):[c in Q => [(a,c) in F => c=b]]]].
> >
> > Note that this does NOT require that ALL(a):ALL(b):[(a, b) in F => a in P & b in Q]. The is does NOT rule out other terms in F.
> >
> Dan Christensen schrieb am Dienstag, 7. Dezember 2021 um 07:00:34 UTC+1:
> > (Correction)
> > On Monday, December 6, 2021 at 2:32:37 AM UTC-5, Mostowski Collapse wrote:
> > > Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
> > >
> > > ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
> > >
> > > Now take contraposition you get:
> > >
> > > ALL(a):ALL(b):[~P(a) | ~Q(b) => ~F(a,b)]
> > >
> > > Or in words, if (a,b) is outside of the domain and co-domain, then F is false.
> > >
> > > https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
> > Rethinking the notion of a function just a bit...
> >
> > Define a set of ordered pairs F such that ALL(a):[a in P => EXIST(b):[b in Q & (a,b) in F & ALL(c):[c in Q => [(a,c) in F => c=b]]]].
> >
> > Note that this does NOT require that ALL(a):ALL(b):[(a, b) in F => a in P & b in Q]. The is does NOT rule out other terms in F.
> >
> For the 1000-th time, thats not the wiki definition.
> Whats wrong with you? Cant you read what is written:
> > Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
> > ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
> https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
>
> It says, a function F : P -> Q is:
> - a binary relation F ⊆ P x Q
> - serial
> - functional
>
There's just a minor technical problem with that approach, as you yourself pointed out: It enables to you assign truth values to functional expressions of the form (a, b) in F outside the domain P. It's a kind of paradox that is resolved as above by tweaking the usual notion of a function as a set of ordered pairs. Or you could just start with the standard functional notation by stating: ALL(a):[a in P => f(a) in Q. That it is my preference.
> Whats wrong with you? Cant you read what is written:
> > Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
> > ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
> https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
>
> It says, a function F : P -> Q is:
> - a binary relation F ⊆ P x Q
> - serial
> - functional
>
Mostowski Collapse
Dec 7, 2021, 12:00:55 PM12/7/21
to
How can a grown up talk so much nonsense!
Its not tweaking the usual notion of a function as a set of ordered pairs.
It is the usual notion of a function as a set of ordered pairs.
Lets get back to F : {0} -> {0}, the only function is F = {(0,0)} a singleton
set of an ordered pair. This implies ~F(0,1) because:
~ (0,1) e {(0,0)}
Try proving ~F(0,1) from your definition, which is not a closed set
of ordered pairs. It leaves pretty open what happens outside
of the domain. You could even have non ordered pair members
in F, outside of the domain?
Its not tweaking the usual notion of a function as a set of ordered pairs.
It is the usual notion of a function as a set of ordered pairs.
Lets get back to F : {0} -> {0}, the only function is F = {(0,0)} a singleton
set of an ordered pair. This implies ~F(0,1) because:
~ (0,1) e {(0,0)}
Try proving ~F(0,1) from your definition, which is not a closed set
of ordered pairs. It leaves pretty open what happens outside
of the domain. You could even have non ordered pair members
in F, outside of the domain?
Mostowski Collapse
Dec 7, 2021, 12:03:28 PM12/7/21
to
Corr.:
This also implies ~F(1,0) because:
~ (1,0) e {(0,0)}
Try proving ~F(1,0) from your "prefered" incorrect definition.
This also implies ~F(1,0) because:
~ (1,0) e {(0,0)}
Try proving ~F(1,0) from your "prefered" incorrect definition.
Mostowski Collapse
Dec 7, 2021, 12:15:50 PM12/7/21
to
If you do your nonsensical ALL(a):[a in P => f(a) in Q] , which
is not the f ⊆ P x Q, with a set f:
/* Dan-O-Matiks Nonsense */
1) ALL(a):[a in P => EXIST(b):[(a,b) e f & b in Q]]
You can not only not prove ~(a,b) e f outside of P x Q, you can
also not assure that outside of P x Q the elements, if they exist,
as you possibly have with your indeterminism,
that they are then ordered pairs, if some exist. This means based
on your prefered incorrect definition 1), you cannot prove:
/* Not Provable */
2) ALL(z):[z e f => EXIST(x):EXIST(y):[z = (x,y)]]
is not the f ⊆ P x Q, with a set f:
/* Dan-O-Matiks Nonsense */
1) ALL(a):[a in P => EXIST(b):[(a,b) e f & b in Q]]
You can not only not prove ~(a,b) e f outside of P x Q, you can
also not assure that outside of P x Q the elements, if they exist,
as you possibly have with your indeterminism,
that they are then ordered pairs, if some exist. This means based
on your prefered incorrect definition 1), you cannot prove:
/* Not Provable */
2) ALL(z):[z e f => EXIST(x):EXIST(y):[z = (x,y)]]
Mostowski Collapse
Dec 7, 2021, 12:24:29 PM12/7/21
to
Here is a proof of ALL(z):[z e f => EXIST(x):EXIST(y):[z = (x,y)]] from f ⊆ P x Q:
/* Exy for x in y, and o(x,y) for (x,y) */
∀z(Ezf → ∃x∃y(z=o(x,y) ∧ (Exp ∧ Eyq))) → ∀z(Ezf → ∃x∃yz=o(x,y)) is valid.
https://www.umsu.de/trees/#~6z%28Ezf~5~7x~7y%28z=o%28x,y%29~1Exp~1Eyq%29%29~5~6z%28Ezf~5~7x~7y%28z=o%28x,y%29%29%29
And here is evidence that it is not provable from ALL(a):[a in P => EXIST(b):[(a,b) e f & b in Q]]:
∀a(Eap → ∃b(Eo(a,b)f ∧ Ebq)) → ∀z(Ezf → ∃x∃yz=o(x,y)) is invalid.
https://www.umsu.de/trees/#~6a%28Eap~5~7b%28Eo%28a,b%29f~1Ebq%29%29~5~6z%28Ezf~5~7x~7y%28z=o%28x,y%29%29%29
/* Exy for x in y, and o(x,y) for (x,y) */
∀z(Ezf → ∃x∃y(z=o(x,y) ∧ (Exp ∧ Eyq))) → ∀z(Ezf → ∃x∃yz=o(x,y)) is valid.
https://www.umsu.de/trees/#~6z%28Ezf~5~7x~7y%28z=o%28x,y%29~1Exp~1Eyq%29%29~5~6z%28Ezf~5~7x~7y%28z=o%28x,y%29%29%29
And here is evidence that it is not provable from ALL(a):[a in P => EXIST(b):[(a,b) e f & b in Q]]:
∀a(Eap → ∃b(Eo(a,b)f ∧ Ebq)) → ∀z(Ezf → ∃x∃yz=o(x,y)) is invalid.
https://www.umsu.de/trees/#~6a%28Eap~5~7b%28Eo%28a,b%29f~1Ebq%29%29~5~6z%28Ezf~5~7x~7y%28z=o%28x,y%29%29%29
Mostowski Collapse
Dec 7, 2021, 12:30:46 PM12/7/21
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Its even worse, when P and Q are the universal class V, then
it also does not assure that f is a set of ordered paris:
∀a∃bEo(a,b)f → ∀z(Ezf → ∃x∃yz=o(x,y)) is invalid.
https://www.umsu.de/trees/#~6a~7bEo%28a,b%29f~5~6z%28Ezf~5~7x~7y%28z=o%28x,y%29%29%29
Your ∀∃ combination is an edifice of imbecility. So there
is something fundamentally wrong in dropping the
assumption f ⊆ P x Q. Its just a ticket to hells kitchen.
it also does not assure that f is a set of ordered paris:
∀a∃bEo(a,b)f → ∀z(Ezf → ∃x∃yz=o(x,y)) is invalid.
https://www.umsu.de/trees/#~6a~7bEo%28a,b%29f~5~6z%28Ezf~5~7x~7y%28z=o%28x,y%29%29%29
Your ∀∃ combination is an edifice of imbecility. So there
is something fundamentally wrong in dropping the
assumption f ⊆ P x Q. Its just a ticket to hells kitchen.
Mostowski Collapse
Dec 7, 2021, 12:38:55 PM12/7/21
to
I wouldn't recommend house painting for Dan-O-Matik.
If you say paint this house, he might also paint the neighbour
house as well, since it was indeterminate what should happen
to the neighbour house. Dan-O-Matik as a painter might
cost you some extra paint and insurance incidents. Dan-O-Matik
with his fluffy function definition just created a new commedy
genre, called anti-occams razor mathematics.
💀 of Math and ☠️ of Physics Archimedes "Putin's Stooge" Plutonium
<plutonium....@gmail.com> blithered:
If you say paint this house, he might also paint the neighbour
house as well, since it was indeterminate what should happen
to the neighbour house. Dan-O-Matik as a painter might
cost you some extra paint and insurance incidents. Dan-O-Matik
with his fluffy function definition just created a new commedy
genre, called anti-occams razor mathematics.
💀 of Math and ☠️ of Physics Archimedes "Putin's Stooge" Plutonium
<plutonium....@gmail.com> blithered:
Dan Christensen
Dec 7, 2021, 1:06:42 PM12/7/21
to
On Tuesday, December 7, 2021 at 12:00:55 PM UTC-5, Mostowski Collapse wrote:
> Dan Christensen schrieb am Dienstag, 7. Dezember 2021 um 16:49:20 UTC+1:
> > There's just a minor technical problem with that approach, as you yourself pointed out: It enables to you assign truth values to functional expressions of the form (a, b) in F outside the domain P. It's a kind of paradox that is resolved as above by tweaking the usual notion of a function as a set of ordered pairs. Or you could just start with the standard functional notation by stating: ALL(a):[a in P => f(a) in Q. That it is my preference.
> Dan Christensen schrieb am Dienstag, 7. Dezember 2021 um 16:49:20 UTC+1:
> > There's just a minor technical problem with that approach, as you yourself pointed out: It enables to you assign truth values to functional expressions of the form (a, b) in F outside the domain P. It's a kind of paradox that is resolved as above by tweaking the usual notion of a function as a set of ordered pairs. Or you could just start with the standard functional notation by stating: ALL(a):[a in P => f(a) in Q. That it is my preference.
> Lets get back to F : {0} -> {0}, the only function is F = {(0,0)} a singleton
> set of an ordered pair. This implies ~F(0,1) because:
>
> ~ (0,1) e {(0,0)}
>
Thanks again for bringing this to our attention, but I see it as problematic -- a paradox that needs to resolved. You probably need an axiom or theorem to the effect that, for all x, there exists a set S such that x in S. Apply it for (0, 0) to obtain (0, 0) in F, which may or may not have elements other than (0, 0). You could not then infer that (x, y) in F => x in {0) & y in {0}. Problem solved!
> set of an ordered pair. This implies ~F(0,1) because:
>
> ~ (0,1) e {(0,0)}
>
Mostowski Collapse
Dec 7, 2021, 2:14:54 PM12/7/21
to
If stupidity had a name, it would be Dan-O-Matik.
> > Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
> > ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
> https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
> It says, a function F : P -> Q is:
> - a binary relation F ⊆ P x Q
> - serial on P,Q
-> functional on P,Q
How many binary relations F ⊆ {0} x {0} are there, that are serial?
> > Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
> > ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
> https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
> It says, a function F : P -> Q is:
> - a binary relation F ⊆ P x Q
-> functional on P,Q
How many binary relations F ⊆ {0} x {0} are there, that are serial?
Dan Christensen
Dec 7, 2021, 2:27:47 PM12/7/21
to
On Tuesday, December 7, 2021 at 2:14:54 PM UTC-5, Mostowski Collapse wrote:
> Dan Christensen schrieb am Dienstag, 7. Dezember 2021 um 19:06:42 UTC+1:
> > On Tuesday, December 7, 2021 at 12:00:55 PM UTC-5, Mostowski Collapse wrote:
> >
> > > Dan Christensen schrieb am Dienstag, 7. Dezember 2021 um 16:49:20 UTC+1:
> > > > There's just a minor technical problem with that approach, as you yourself pointed out: It enables to you assign truth values to functional expressions of the form (a, b) in F outside the domain P. It's a kind of paradox that is resolved as above by tweaking the usual notion of a function as a set of ordered pairs. Or you could just start with the standard functional notation by stating: ALL(a):[a in P => f(a) in Q. That it is my preference.
> > > Lets get back to F : {0} -> {0}, the only function is F = {(0,0)} a singleton
> > > set of an ordered pair. This implies ~F(0,1) because:
> > >
> > > ~ (0,1) e {(0,0)}
> > >
> >
> > Thanks again for bringing this to our attention, but I see it as problematic -- a paradox that needs to [be] resolved. You probably need an axiom or theorem to the effect that, for all x, there exists a set S such that x in S [other elements not being ruled out] . Apply it for (0, 0) to obtain (0, 0) in F, which may or may not have elements other than (0, 0). You could not then infer that (x, y) in F => x in {0) & y in {0}. Problem solved!
> Dan Christensen schrieb am Dienstag, 7. Dezember 2021 um 19:06:42 UTC+1:
> > On Tuesday, December 7, 2021 at 12:00:55 PM UTC-5, Mostowski Collapse wrote:
> >
> > > Dan Christensen schrieb am Dienstag, 7. Dezember 2021 um 16:49:20 UTC+1:
> > > > There's just a minor technical problem with that approach, as you yourself pointed out: It enables to you assign truth values to functional expressions of the form (a, b) in F outside the domain P. It's a kind of paradox that is resolved as above by tweaking the usual notion of a function as a set of ordered pairs. Or you could just start with the standard functional notation by stating: ALL(a):[a in P => f(a) in Q. That it is my preference.
> > > Lets get back to F : {0} -> {0}, the only function is F = {(0,0)} a singleton
> > > set of an ordered pair. This implies ~F(0,1) because:
> > >
> > > ~ (0,1) e {(0,0)}
> > >
> >
> > > Check wikipedia, F : P -> Q, requires necessarely F ⊆ P x Q:
> > > ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)]
> > https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
>
> > It says, a function F : P -> Q is:
> > - a binary relation F ⊆ P x Q
> > - serial on P,Q
> -> functional on P,Q
>
Mostowski Collapse
Dec 7, 2021, 2:33:32 PM12/7/21
to
Depending how you do the ordered pairs in F ⊆ P x Q ,
yes you might need the pairing axiom of ZFC. Even
the separation axiom might be needed, and the power
axiom of course as well. But I don't think that you
need it when you use the class based formulation
like here, ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)] , where
P,Q and F are predicate symbols, P and Q unary,
F binary. What do you want Dan-O-Matik. I asked
for beginner math, and said it needs set theory. You
came up with the class based formulation which
avoids any set theory. It doesn't change anything that
you always sweep F ⊆ P x Q under the rug.
Because:
> 1) Wikipedia definition of a function:
> a binary relation F ⊆ P x Q, serial on P,Q and functional on P,Q
>
> 2) Dan-O-Matiks psychotic definition:
> serial on P,Q and functional on P,Q
Your ∀∃ combination is only the serial part. Read
wikipedia, its right in front of you:
A binary relation is serial (also called left-total) if
∀ x ∈ X , ∃ y ∈ Y , ( x , y ) ∈ R .
https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
Its the same as:
∀x(x ∈ X => ∃y(y ∈ Y & ( x , y ) ∈ R))
Or reordering the conjunction:
∀x(x ∈ X => ∃y( ( x , y ) ∈ R & y ∈ Y))
If you have function application the following might be tolerated:
∀x(x ∈ X => R(x) ∈ Y))
But you need a logic different from FOL to deal correctly
with R(x) in general.
yes you might need the pairing axiom of ZFC. Even
the separation axiom might be needed, and the power
axiom of course as well. But I don't think that you
need it when you use the class based formulation
like here, ALL(a):ALL(b):[F(a,b) => P(a) & Q(b)] , where
P,Q and F are predicate symbols, P and Q unary,
F binary. What do you want Dan-O-Matik. I asked
for beginner math, and said it needs set theory. You
came up with the class based formulation which
avoids any set theory. It doesn't change anything that
you always sweep F ⊆ P x Q under the rug.
Because:
> 1) Wikipedia definition of a function:
> a binary relation F ⊆ P x Q, serial on P,Q and functional on P,Q
>
> 2) Dan-O-Matiks psychotic definition:
> serial on P,Q and functional on P,Q
wikipedia, its right in front of you:
A binary relation is serial (also called left-total) if
∀ x ∈ X , ∃ y ∈ Y , ( x , y ) ∈ R .
https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
Its the same as:
∀x(x ∈ X => ∃y(y ∈ Y & ( x , y ) ∈ R))
Or reordering the conjunction:
∀x(x ∈ X => ∃y( ( x , y ) ∈ R & y ∈ Y))
If you have function application the following might be tolerated:
∀x(x ∈ X => R(x) ∈ Y))
But you need a logic different from FOL to deal correctly
with R(x) in general.
Mostowski Collapse
Dec 7, 2021, 2:50:44 PM12/7/21
to
There is no paradox and you didn't propose something.
You only used serial in the past, recently you used serial+
functional as well, you had two variants of serial+functional,
but you always sweep under the rug binary relation.
LoL
You only used serial in the past, recently you used serial+
functional as well, you had two variants of serial+functional,
but you always sweep under the rug binary relation.
LoL
Dan Christensen
Dec 7, 2021, 3:22:33 PM12/7/21
to
On Tuesday, December 7, 2021 at 2:33:32 PM UTC-5, Mostowski Collapse wrote:
> Depending how you do the ordered pairs in F ⊆ P x Q ,
> yes you might need the pairing axiom of ZFC. Even
> the separation axiom might be needed, and the power
> axiom of course as well.
For functions expressed as a set of ordered pairs, with domain p and codomain q:
> Depending how you do the ordered pairs in F ⊆ P x Q ,
> yes you might need the pairing axiom of ZFC. Even
> the separation axiom might be needed, and the power
> axiom of course as well.
ALL(f):ALL(p):ALL(q):[Set'(f) & Set(p) & Set(q)
=> ALL(a):[a in p => EXIST(b):[b in q & (a,b) in f & ALL(c):[c in q => [(a,c) in f => c=b]]]]]
Since it cannot be ruled, for example, that there may be an ordered-pair (x, y) in f such that x not in p, you won't run into the paradoxical situation that you pointed out.
Mostowski Collapse
Dec 7, 2021, 4:25:15 PM12/7/21
to
I didn't point to some paradox.
But what you posted is still non-sense, its only serial+functional.
Try p={0} and q={0}. With your definition you cannot prove:
~(1,0) e f
So its not the correct definiton of functions as set of ordered pairs.
But what you posted is still non-sense, its only serial+functional.
Try p={0} and q={0}. With your definition you cannot prove:
~(1,0) e f
So its not the correct definiton of functions as set of ordered pairs.
Mostowski Collapse
Dec 7, 2021, 4:29:00 PM12/7/21
to
Could it be that you dont understand logic, in particular material implication?
Mostowski Collapse
Dec 7, 2021, 4:34:12 PM12/7/21
to
Your error is still that you cannot prove: ~(1,0) e f ,
since ~1 e {0} your definition becomes
VACOUSLY TRUE
And hence your defintion cannot exclude (1,0) e f.
Your definition is only serial+functional.
So what is missing? Same old story:
> STUDENTS BEWARE Dan-O-Matik didn't take his medication,
> sweeping under the rug parts of a definitions. Lets recap:
>
since ~1 e {0} your definition becomes
VACOUSLY TRUE
And hence your defintion cannot exclude (1,0) e f.
Your definition is only serial+functional.
So what is missing? Same old story:
> STUDENTS BEWARE Dan-O-Matik didn't take his medication,
> sweeping under the rug parts of a definitions. Lets recap:
>
> 1) Wikipedia definition of a function:
> a binary relation F ⊆ P x Q, serial on P,Q and functional on P,Q
>
> 2) Dan-O-Matiks psychotic definition:
> serial on P,Q and functional on P,Q
>
> a binary relation F ⊆ P x Q, serial on P,Q and functional on P,Q
>
> 2) Dan-O-Matiks psychotic definition:
> serial on P,Q and functional on P,Q
>
> Do you see any differents dear students?
Dan Christensen
Dec 7, 2021, 5:00:11 PM12/7/21
to
On Tuesday, December 7, 2021 at 4:25:15 PM UTC-5, Mostowski Collapse wrote:
> Dan Christensen schrieb am Dienstag, 7. Dezember 2021 um 21:22:33 UTC+1:
> > On Tuesday, December 7, 2021 at 2:33:32 PM UTC-5, Mostowski Collapse wrote:
> > > Depending how you do the ordered pairs in F ⊆ P x Q ,
> > > yes you might need the pairing axiom of ZFC. Even
> > > the separation axiom might be needed, and the power
> > > axiom of course as well.
> > For functions expressed as a set of ordered pairs, with domain p and codomain q:
> >
> > ALL(f):ALL(p):ALL(q):[Set'(f) & Set(p) & Set(q)
> > => ALL(a):[a in p => EXIST(b):[b in q & (a,b) in f & ALL(c):[c in q => [(a,c) in f => c=b]]]]]
> >
> > Since it cannot be ruled [out], for example, that there may be an ordered-pair (x, y) in f such that x not in p, you won't run into the paradoxical situation that you pointed out.
> Dan Christensen schrieb am Dienstag, 7. Dezember 2021 um 21:22:33 UTC+1:
> > On Tuesday, December 7, 2021 at 2:33:32 PM UTC-5, Mostowski Collapse wrote:
> > > Depending how you do the ordered pairs in F ⊆ P x Q ,
> > > yes you might need the pairing axiom of ZFC. Even
> > > the separation axiom might be needed, and the power
> > > axiom of course as well.
> > For functions expressed as a set of ordered pairs, with domain p and codomain q:
> >
> > ALL(f):ALL(p):ALL(q):[Set'(f) & Set(p) & Set(q)
> > => ALL(a):[a in p => EXIST(b):[b in q & (a,b) in f & ALL(c):[c in q => [(a,c) in f => c=b]]]]]
> >
> I didn't point to some paradox.
>
> But what you posted is still non-sense, its only serial+functional.
>
> Try p={0} and q={0}. With your definition you cannot prove:
>
> ~(1,0) e f
>
> So its not the correct definiton of functions as set of ordered pairs.
Mostowski Collapse
Dec 7, 2021, 7:17:46 PM12/7/21
to
There is only one mapping F : {0} -> {0}, it is:
F = {(0,0)}
Therefore ~(1,0) e F. You cannot discuss this away.
F = {(0,0)}
Therefore ~(1,0) e F. You cannot discuss this away.
Dan Christensen
Dec 7, 2021, 7:52:31 PM12/7/21
to
On Tuesday, December 7, 2021 at 7:17:46 PM UTC-5, Mostowski Collapse wrote:
> Dan Christensen schrieb am Dienstag, 7. Dezember 2021 um 23:00:11 UTC+1:
> > > Try p={0} and q={0}. With your definition you cannot prove:
> > >
> > > ~(1,0) e f
> > >
> > Of course not. I would be worried if I could. You should be worried that you CAN prove it.
> > > So its not the correct definiton of functions as set of ordered pairs.
> > From your example, we see that whatever "definition" you may be using is not always correct.
> Dan Christensen schrieb am Dienstag, 7. Dezember 2021 um 23:00:11 UTC+1:
> > > Try p={0} and q={0}. With your definition you cannot prove:
> > >
> > > ~(1,0) e f
> > >
> > Of course not. I would be worried if I could. You should be worried that you CAN prove it.
> > > So its not the correct definiton of functions as set of ordered pairs.
> > From your example, we see that whatever "definition" you may be using is not always correct.
> There is only one mapping F : {0} -> {0}, it is:
>
> F = {(0,0)}
>
In functional notation, that would be f(0)=0.
>
> F = {(0,0)}
>
While they may suffice in informal mathematics, as we have seen with your example, such a set of ordered pairs may the not the best representation of a function in rigorous mathematics.
> Therefore ~(1,0) e F. You cannot discuss this away.
Reese Page
Dec 7, 2021, 7:55:56 PM12/7/21
to
Mostowski Collapse wrote:
> If stupidity had a name, it would be Dan-O-Matik.
stupidity has no name??
> If stupidity had a name, it would be Dan-O-Matik.
FromTheRafters
Dec 8, 2021, 5:48:48 AM12/8/21
to
Dan Christensen brought next idea :
domain "Kings of France" and as such cannot have a bald/hairy truth
value other than undefined.
> On Tuesday, December 7, 2021 at 4:25:15 PM UTC-5, Mostowski Collapse wrote:
>
>
>> Dan Christensen schrieb am Dienstag, 7. Dezember 2021 um 21:22:33 UTC+1:
>>> On Tuesday, December 7, 2021 at 2:33:32 PM UTC-5, Mostowski Collapse wrote:
>>>> Depending how you do the ordered pairs in F ⊆ P x Q ,
>>>> yes you might need the pairing axiom of ZFC. Even
>>>> the separation axiom might be needed, and the power
>>>> axiom of course as well.
>>> For functions expressed as a set of ordered pairs, with domain p and
>>> codomain q:
>>>
>>> ALL(f):ALL(p):ALL(q):[Set'(f) & Set(p) & Set(q)
>>> => ALL(a):[a in p => EXIST(b):[b in q & (a,b) in f & ALL(c):[c in q =>
>>> [(a,c) in f => c=b]]]]]
>>>
>>> Since it cannot be ruled [out], for example, that there may be an
>>> ordered-pair (x, y) in f such that x not in p, you won't run into the
>>> paradoxical situation that you pointed out.
>
>> I didn't point to some paradox.
>>
>
> On the contrary, attaching to a truth-value to expressions like f(x)=y for x
> not in the domain of f is very paradoxical. Common sense should tell you that
> its truth value is undefined.
Seems similar to the idea that a present king of France is not in the
>
>
>> Dan Christensen schrieb am Dienstag, 7. Dezember 2021 um 21:22:33 UTC+1:
>>> On Tuesday, December 7, 2021 at 2:33:32 PM UTC-5, Mostowski Collapse wrote:
>>>> Depending how you do the ordered pairs in F ⊆ P x Q ,
>>>> yes you might need the pairing axiom of ZFC. Even
>>>> the separation axiom might be needed, and the power
>>>> axiom of course as well.
>>> For functions expressed as a set of ordered pairs, with domain p and
>>> codomain q:
>>>
>>> ALL(f):ALL(p):ALL(q):[Set'(f) & Set(p) & Set(q)
>>> => ALL(a):[a in p => EXIST(b):[b in q & (a,b) in f & ALL(c):[c in q =>
>>> [(a,c) in f => c=b]]]]]
>>>
>>> Since it cannot be ruled [out], for example, that there may be an
>>> ordered-pair (x, y) in f such that x not in p, you won't run into the
>>> paradoxical situation that you pointed out.
>
>> I didn't point to some paradox.
>>
>
> On the contrary, attaching to a truth-value to expressions like f(x)=y for x
> not in the domain of f is very paradoxical. Common sense should tell you that
> its truth value is undefined.
domain "Kings of France" and as such cannot have a bald/hairy truth
value other than undefined.
Message has been deleted
Message has been deleted
Dan Christensen
Dec 8, 2021, 12:48:28 PM12/8/21
to
(Correction 2)
We not speculating here about the attributes of an imaginary being. The difference is that Jan Burse (MC) is talking about elements actually existing outside the domain of a function.
Burse's Paradox: If f: A --> B then f(x)=/=y for all x not in A.
This is a result of thinking of f as a subset of AxB, a common enough practice, but it does lead to this wonky result.
One of the essential features of a function is that it is defined ONLY for elements of its domain. f(x) for x not in the domain of f is quite meaningless. It should not be used to make logical inferences as above.
To resolve this paradox, it may be useful to think of a function f with domain A and codomain B as a set that is not explicitly restricted to elements of AxB as follows:
ALL(x):[x in A => EXIST(y):[y in B & (x,y) in f & ALL(z):[z in B => [(x,z) in f => z=y]]]]
This does not explicitly exclude any "junk terms" from the set f, thus we avoid Burse's Paradox.
To introduce an arbitrary function f: A --> B in a proof, one need only state:
ALL(x):[x in A => f(x) in B]
This definition cannot be applied for x not in A, thus avoiding BP.
Burse's Paradox: If f: A --> B then f(x)=/=y for all x not in A.
This is a result of thinking of f as a subset of AxB, a common enough practice, but it does lead to this wonky result.
One of the essential features of a function is that it is defined ONLY for elements of its domain. f(x) for x not in the domain of f is quite meaningless. It should not be used to make logical inferences as above.
To resolve this paradox, it may be useful to think of a function f with domain A and codomain B as a set that is not explicitly restricted to elements of AxB as follows:
ALL(x):[x in A => EXIST(y):[y in B & (x,y) in f & ALL(z):[z in B => [(x,z) in f => z=y]]]]
This does not explicitly exclude any "junk terms" from the set f, thus we avoid Burse's Paradox.
To introduce an arbitrary function f: A --> B in a proof, one need only state:
ALL(x):[x in A => f(x) in B]
This definition cannot be applied for x not in A, thus avoiding BP.
Jim Burns
Dec 8, 2021, 2:57:55 PM12/8/21
to
On 12/8/2021 12:48 PM, Dan Christensen wrote:
> On Wednesday, December 8, 2021 at 5:48:48 AM UTC-5,
> FromTheRafters wrote:
>> Seems similar to the idea that a present king of France
>> is not in the domain "Kings of France" and as such cannot
>> have a bald/hairy truth value other than undefined.
>
> We not speculating here about the attributes of an imaginary
> being.
The present king of France is _non-existent_
> On Wednesday, December 8, 2021 at 5:48:48 AM UTC-5,
> FromTheRafters wrote:
>> Seems similar to the idea that a present king of France
>> is not in the domain "Kings of France" and as such cannot
>> have a bald/hairy truth value other than undefined.
>
> We not speculating here about the attributes of an imaginary
> being.
We can and do reason about imaginary beings.
Whether it matters what our conclusions are is
a whole 'nother can of worms, but we do.
I think those conclusions matter.
Think of (actual, no-joke) dark matter as an imaginary
being. Or luminiferous aether. Or elan vital.
We need to be able to reason about these entities
which we imagine, whether they exist or not, in order to
decide whether they exist.
I would say the distinction between imaginary and
non-existent winds through the foundations of our reasoning.
> The difference is that Jan Burse (MC) is talking about
> elements actually existing outside the domain of a function.
>
> Burse's Paradox:
> If f: A --> B then f(x)=/=y for all x not in A.
It's a matter of definition.
We are characterizing a function *from* A *to* B
Consider the factorial k!
∀j ∈ ℕ, ∃!k ∈ ℕ, k = j!
The gamma function extends the factorial to almost
all of the complex plane ℂ
ℂ⁺ == ℂ\{0,-1,-2,-3,...}
∀z ∈ ℂ⁺, ∃!w ∈ ℂ, w = Γ(z)
∀j ∈ ℕ, Γ(j) = (j-1)!
so gamma and factorial are closely related.
But we have defined "function" so that they are
different functions.
Maybe you would like "function" defined other than
how it is, but that is how it is defined.
We have our reasons for doing so, and you might well
be able to work out what they are. You might even
ask someone. But it's a definition, not a paradox.
Dan Christensen
Dec 8, 2021, 4:17:30 PM12/8/21
to
See my posting in this topic at MSE:
https://math.stackexchange.com/questions/4325129/what-is-the-truth-value-if-any-for-fx-y-when-x-is-outside-of-the-domain-of/4327829#4327829
Dan
https://math.stackexchange.com/questions/4325129/what-is-the-truth-value-if-any-for-fx-y-when-x-is-outside-of-the-domain-of/4327829#4327829
Dan
Dan Christensen
Dec 8, 2021, 10:20:12 PM12/8/21
to
On Wednesday, December 8, 2021 at 2:57:55 PM UTC-5, Jim Burns wrote:
>
> Maybe you would like "function" defined other than
> how it is, but that is how it is defined.
It's hard to imagine that some important result would hinge on what the value of a function outside its domain is not, so this is probably just a a tempest in a teacup. But if someone needs a formal definition that will ensure that a function's value remains undefined outside its domain, this may suit their purposes.
>
> Maybe you would like "function" defined other than
> how it is, but that is how it is defined.
Dan Christensen
Dec 8, 2021, 10:24:59 PM12/8/21
to
On Wednesday, December 8, 2021 at 10:20:12 PM UTC-5, Dan Christensen wrote:
> On Wednesday, December 8, 2021 at 2:57:55 PM UTC-5, Jim Burns wrote:
>
> >
> > Maybe you would like "function" defined other than
> > how it is, but that is how it is defined.
> It's hard to imagine that some important result would hinge on what the value of a function outside its domain is not, so this is probably just a a tempest in a teacup. But if someone needs a formal definition that will ensure that a function's value remains undefined outside its domain, this may suit their purposes.
ALL(x):[x in A => f(x) in B] would also work.
> On Wednesday, December 8, 2021 at 2:57:55 PM UTC-5, Jim Burns wrote:
>
> >
> > Maybe you would like "function" defined other than
> > how it is, but that is how it is defined.
> It's hard to imagine that some important result would hinge on what the value of a function outside its domain is not, so this is probably just a a tempest in a teacup. But if someone needs a formal definition that will ensure that a function's value remains undefined outside its domain, this may suit their purposes.
Dan Christensen
Dec 9, 2021, 5:19:25 PM12/9/21
to
Mostowski Collapse
Dec 9, 2021, 6:53:27 PM12/9/21
to
Its very easy to prove that (x,y) e f is false outside of the domain.
But you cannot prove that f(x)=y is false outside of the domain.
Because in FOL the f in f(x) denotes a function f : V -> V, a second
order object, whereas the f in (x,y) e f denotes an element f : V a
first order object. The problem with your axiom:
ALL(x):[x in A => f(x) in B]
It talks about a second order object f, which is anyway f : V -> V.
You cannot coerce it to a smaller domain. You can always
prove the following:
ALL(x):EXIST(y):f(x)=y
If you would add a binary relation axiom, as per wiki f ⊆ P x Q,
but in wiki f is a first order object or a class, but not a second order
object. If you add this axiom:
(**) ALL(x):ALL(y):[f(x)=y -> P(x) & Q(y)]
You get an inconsistent system. Try adding (**), you will be
able to derive false in DC Proof. Thats relatively simple.
But you cannot prove that f(x)=y is false outside of the domain.
Because in FOL the f in f(x) denotes a function f : V -> V, a second
order object, whereas the f in (x,y) e f denotes an element f : V a
first order object. The problem with your axiom:
ALL(x):[x in A => f(x) in B]
You cannot coerce it to a smaller domain. You can always
prove the following:
ALL(x):EXIST(y):f(x)=y
If you would add a binary relation axiom, as per wiki f ⊆ P x Q,
but in wiki f is a first order object or a class, but not a second order
object. If you add this axiom:
(**) ALL(x):ALL(y):[f(x)=y -> P(x) & Q(y)]
You get an inconsistent system. Try adding (**), you will be
able to derive false in DC Proof. Thats relatively simple.
Mostowski Collapse
Dec 9, 2021, 7:02:17 PM12/9/21
to
Here is the inconsistency proved:
(∃x¬Px ∧ ∀x∀y(f(x)=y → (Px ∧ Qy))) → (R∧¬R) is valid.
https://www.umsu.de/trees/#~7x~3Px~1~6x~6y%28f%28x%29=y~5Px~1Qy%29~5R~1~3R
In ZFC if P is set-like, you can always prove ∃x¬Px.
And any class P different from V, as also a separating element.
(∃x¬Px ∧ ∀x∀y(f(x)=y → (Px ∧ Qy))) → (R∧¬R) is valid.
https://www.umsu.de/trees/#~7x~3Px~1~6x~6y%28f%28x%29=y~5Px~1Qy%29~5R~1~3R
In ZFC if P is set-like, you can always prove ∃x¬Px.
And any class P different from V, as also a separating element.
Message has been deleted
Mostowski Collapse
Dec 9, 2021, 7:39:24 PM12/9/21
to
Pitty, I guess, the tree tool has a bug:
∀x∀y(f(x)=y → (Px ∧ Qy)) → (R∧¬R) is invalid.
https://www.umsu.de/trees/#~6x~6y%28f%28x%29=y~5Px~1Qy%29~5R~1~3R
Domain: | { 0 }
-- | --
f: | { }
P: | { 0 }
Q: | { 0 }
R: | false
It seems to me that the counter model is not a first oder model.
What is an empty function f = {} ? Looks rather like a partial function
to me. On the other hand the same system can prove the following,
saying f is total:
∀x∃yf(x)=y is valid.
https://www.umsu.de/trees/#~6x~7yf%28x%29=y
∀x∀y(f(x)=y → (Px ∧ Qy)) → (R∧¬R) is invalid.
https://www.umsu.de/trees/#~6x~6y%28f%28x%29=y~5Px~1Qy%29~5R~1~3R
Domain: | { 0 }
-- | --
f: | { }
P: | { 0 }
Q: | { 0 }
R: | false
It seems to me that the counter model is not a first oder model.
What is an empty function f = {} ? Looks rather like a partial function
to me. On the other hand the same system can prove the following,
saying f is total:
∀x∃yf(x)=y is valid.
https://www.umsu.de/trees/#~6x~7yf%28x%29=y
Dan Christensen
Dec 9, 2021, 10:30:25 PM12/9/21
to
On Thursday, December 9, 2021 at 6:53:27 PM UTC-5, Mostowski Collapse wrote:
> Its very easy to prove that (x,y) e f is false outside of the domain.
I thinks it's called Burse's Paradox. You really must find some way around it, Jan Burse. Somehow, you must go from a set of ordered pairs to an actual function which is undefined outside of its domain set. Here is how it is done in DC Proof: For functions of one variable, we have an axiom (from the Sets menu):
> Its very easy to prove that (x,y) e f is false outside of the domain.
1 ALL(f):ALL(dom):ALL(cod):[Set'(f) & Set(dom) & Set(cod)
=> [ALL(a1):ALL(b):[(a1,b) in f => a1 in dom & b in cod]
& ALL(a1):[a1 in dom => EXIST(b):[b in cod & (a1,b) in f]]
& ALL(a1):ALL(b1):ALL(b2):[a1 in dom & b1 in cod & b2 in cod
=> [(a1,b1) in f & (a1,b2) in f => b1=b2]]
=> ALL(a1):ALL(b):[a1 in dom & b in cod => [f(a1)=b <=> (a1,b) in f]]]]
Function
You would use this axiom to construct, i.e. prove the existence of a particular function. You start with the sets f, dom and cod, prove they have certain properties expected of a function. Then you will be able to prove:
ALL(x):[x in dom => f(x) in cod]
Sorry, but you will NOT be able to prove f(x)=/=y for any x not in dom. You will not be able to infer anything about f(x) if x is not in the domain of f. Just like in almost every math textbook on the planet.
If you just want to introduce an arbitrary function, you can skip all the above and simply state from the start:
ALL(x):[x in dom => f(x) in cod]
As above, you will not be able to make any inference about f(x) if x is not in the domain of f.
I hope this helps you deal with your paradox, Jan Burse.
Mostowski Collapse
Dec 10, 2021, 9:21:53 AM12/10/21
to
You can call it whatever you want. You can also
snip of my posts whatever you want.
But I wrote:
> Its very easy to prove that (x,y) e f is false outside of the domain.
> But you cannot prove that f(x)=y is false outside of the domain.
Try adding these axioms:
1) ∃x¬Px
P is not the universal class.
2) ∀x∀y(f(x)=y → (Px ∧ Qy))
f is a binary relation
You can derive an inconsistency. You should be also able to dervie
an inconistency in DC Proof. Should I do it for you?
BTW: Meanwhile Wolfgang Schwartz fixed his model finder,
it doesn't show the f={} anomaly anymore.
https://github.com/wo/tpg/issues/8
snip of my posts whatever you want.
But I wrote:
> Its very easy to prove that (x,y) e f is false outside of the domain.
Try adding these axioms:
1) ∃x¬Px
P is not the universal class.
2) ∀x∀y(f(x)=y → (Px ∧ Qy))
f is a binary relation
You can derive an inconsistency. You should be also able to dervie
an inconistency in DC Proof. Should I do it for you?
BTW: Meanwhile Wolfgang Schwartz fixed his model finder,
it doesn't show the f={} anomaly anymore.
https://github.com/wo/tpg/issues/8
Mostowski Collapse
Dec 10, 2021, 9:30:18 AM12/10/21
to
Proving inconsistency means deriving false from 1) and 2).
You can derive false from 1) and 2):
(∃x¬Px ∧ ∀x∀y(f(x)=y → (Px ∧ Qy))) → (R∧¬R) is valid.
https://www.umsu.de/trees/#~7x~3Px~1~6x~6y%28f%28x%29=y~5Px~1Qy%29~5R~1~3R
But since in classical logic ~A and A -> f are the same,
we can also prove the following:
¬(∃x¬Px ∧ ∀x∀y(f(x)=y → (Px ∧ Qy))) is valid.
https://www.umsu.de/trees/#~3%28~7x~3Px~1~6x~6y%28f%28x%29=y~5Px~1Qy%29%29
Its only 9 proof lines. How many lines will it be in DC Proof?
You can derive false from 1) and 2):
(∃x¬Px ∧ ∀x∀y(f(x)=y → (Px ∧ Qy))) → (R∧¬R) is valid.
https://www.umsu.de/trees/#~7x~3Px~1~6x~6y%28f%28x%29=y~5Px~1Qy%29~5R~1~3R
we can also prove the following:
¬(∃x¬Px ∧ ∀x∀y(f(x)=y → (Px ∧ Qy))) is valid.
https://www.umsu.de/trees/#~3%28~7x~3Px~1~6x~6y%28f%28x%29=y~5Px~1Qy%29%29
Its only 9 proof lines. How many lines will it be in DC Proof?
Mostowski Collapse
Dec 10, 2021, 10:36:48 AM12/10/21
to
Thats a nice strenghening:
∀x(Px → ∃y(Qy ∧ f(x)=y)) → (∀x∀y(f(x)=y → (Px ∧ Qy)) ↔ ∀xPx) is valid.
https://www.umsu.de/trees/#~6x%28Px~5~7y%28Qy~1f%28x%29=y%29%29~5%28~6x~6y%28f%28x%29=y~5Px~1Qy%29~4~6xPx%29
LoL
∀x(Px → ∃y(Qy ∧ f(x)=y)) → (∀x∀y(f(x)=y → (Px ∧ Qy)) ↔ ∀xPx) is valid.
https://www.umsu.de/trees/#~6x%28Px~5~7y%28Qy~1f%28x%29=y%29%29~5%28~6x~6y%28f%28x%29=y~5Px~1Qy%29~4~6xPx%29
LoL
Dan Christensen
Dec 10, 2021, 10:44:21 AM12/10/21
to
On Friday, December 10, 2021 at 9:21:53 AM UTC-5, Mostowski Collapse wrote:
> You can call it whatever you want. You can also
> snip of my posts whatever you want.
> But I wrote:
> > Its very easy to prove that (x,y) e f is false outside of the domain.
> > But you cannot prove that f(x)=y is false outside of the domain.
>
> Try adding these axioms:
>
> 1) ∃x¬Px
> P is not the universal class.
>
Every set excludes something since there is no universal set of all things.
> You can call it whatever you want. You can also
> snip of my posts whatever you want.
> But I wrote:
> > Its very easy to prove that (x,y) e f is false outside of the domain.
> > But you cannot prove that f(x)=y is false outside of the domain.
>
> Try adding these axioms:
>
> 1) ∃x¬Px
> P is not the universal class.
>
> 2) ∀x∀y(f(x)=y → (Px ∧ Qy))
> f is a binary relation
>
> You can derive an inconsistency. You should be also able to dervie
> an inconistency in DC Proof. Should I do it for you?
>
Mostowski Collapse
Dec 10, 2021, 1:07:48 PM12/10/21
to
Dan-O-Matik halucinated:
The strenghening of the inconsistency is:
> ∀x(Px → ∃y(Qy ∧ f(x)=y)) → (∀x∀y(f(x)=y → (Px ∧ Qy)) ↔ ∀xPx) is valid.
https://www.umsu.de/trees/#~6x%28Px~5~7y%28Qy~1f%28x%29=y%29%29~5%28~6x~6y%28f%28x%29=y~5Px~1Qy%29~4~6xPx%29
∀x(Px → ∃y(Qy ∧ f(x)=y)) is "serial", Dan-O-Matiks favorite
∀x∀y(f(x)=y → (Px ∧ Qy)) is "binary relation",
∀xPx is universal class
For "serial" and "binary relation" see wiki:
https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
> "f(x)" makes no sense if you have not specified a domain for x.
I have specified a domain, it is P. What is wrong with you?
The strenghening of the inconsistency is:
> ∀x(Px → ∃y(Qy ∧ f(x)=y)) → (∀x∀y(f(x)=y → (Px ∧ Qy)) ↔ ∀xPx) is valid.
https://www.umsu.de/trees/#~6x%28Px~5~7y%28Qy~1f%28x%29=y%29%29~5%28~6x~6y%28f%28x%29=y~5Px~1Qy%29~4~6xPx%29
∀x∀y(f(x)=y → (Px ∧ Qy)) is "binary relation",
∀xPx is universal class
For "serial" and "binary relation" see wiki:
https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
Mostowski Collapse
Dec 10, 2021, 1:11:55 PM12/10/21
to
Works also like this, simplifying ∀x(Px → ∃y(Qy ∧ f(x)=y))
into ∀x(Px → Qf(x)), same inconsistency:
∀x(Px → Qf(x)) → (∀x∀y(f(x)=y → (Px ∧ Qy)) ↔ ∀xPx) is valid.
https://www.umsu.de/trees/#~6x%28Px~5Qf%28x%29%29~5%28~6x~6y%28f%28x%29=y~5Px~1Qy%29~4~6xPx%29
into ∀x(Px → Qf(x)), same inconsistency:
∀x(Px → Qf(x)) → (∀x∀y(f(x)=y → (Px ∧ Qy)) ↔ ∀xPx) is valid.
https://www.umsu.de/trees/#~6x%28Px~5Qf%28x%29%29~5%28~6x~6y%28f%28x%29=y~5Px~1Qy%29~4~6xPx%29
Mostowski Collapse
Dec 10, 2021, 1:14:46 PM12/10/21
to
Or in words (Dan-O-Matik function = serial function):
A Dan-O-Matik function is a binary relation iff the domain is the universal class
A Dan-O-Matik function is a binary relation iff the domain is the universal class
Dan Christensen
Dec 10, 2021, 5:28:27 PM12/10/21
to
Dan
Mostowski Collapse
Dec 10, 2021, 6:29:21 PM12/10/21
to
I nowhere wrote inconsistency "in DC Proof" in this thread.
Whats wrong with you? What I wrote is this:
> Try adding these axioms:
>
> 1) ∃x¬Px
> P is not the universal class.
>
∃x¬Px, ∀x∀y(f(x)=y → (Px ∧ Qy)) entails R∧¬R.
https://www.umsu.de/trees/#~7x~3Px,~6x~6y%28f%28x%29=y~5Px~1Qy%29|=R~1~3R
Whats wrong with you? What I wrote is this:
> Try adding these axioms:
>
> 1) ∃x¬Px
> P is not the universal class.
>
> 2) ∀x∀y(f(x)=y → (Px ∧ Qy))
> f is a binary relation
>
> f is a binary relation
>
> You can derive an inconsistency.
The tree tool can do it:
∃x¬Px, ∀x∀y(f(x)=y → (Px ∧ Qy)) entails R∧¬R.
https://www.umsu.de/trees/#~7x~3Px,~6x~6y%28f%28x%29=y~5Px~1Qy%29|=R~1~3R
Mostowski Collapse
Dec 10, 2021, 6:40:44 PM12/10/21
to
But you can also add an axiom, as you often do,
in your blog post proof:
(*) ∀x(Px → Qf(x))
And then derive from it:
(**) (∀x∀y(f(x)=y → (Px ∧ Qy)) ↔ ∀xPx)
I called it also "inconsisteny", because its a similar
idea. But the statement us now:
A Dan-O-Matik function is a binary relation iff the domain is the universal class.
in your blog post proof:
(*) ∀x(Px → Qf(x))
And then derive from it:
(**) (∀x∀y(f(x)=y → (Px ∧ Qy)) ↔ ∀xPx)
I called it also "inconsisteny", because its a similar
idea. But the statement us now:
A Dan-O-Matik function is a binary relation iff the domain is the universal class.
Mostowski Collapse
Dec 10, 2021, 6:56:58 PM12/10/21
to
This probably explains why many HOL systems are also
type systems. A HOL system allows quantifying second order
objects, such as functions. So like in DC Proof, in a HOL system
one can do ALL(f) or EXIST(f). One explanation why one has
type systems, is then to avoid paradoxes of self application.
But another explanation could also be that the
functions subject to ALL(f) or EXIST(f) are a little inflexible,
if they are borrowd from FOL, i.e. if the HOL is extension of
FOL, then these function symbols are f : V -> V, and one
cannot coerce them to smaller domains and codomains,
as the inconsistency shows. On the other hand a HOL system
with product types would have types such as function space A -> B.
type systems. A HOL system allows quantifying second order
objects, such as functions. So like in DC Proof, in a HOL system
one can do ALL(f) or EXIST(f). One explanation why one has
type systems, is then to avoid paradoxes of self application.
But another explanation could also be that the
functions subject to ALL(f) or EXIST(f) are a little inflexible,
if they are borrowd from FOL, i.e. if the HOL is extension of
FOL, then these function symbols are f : V -> V, and one
cannot coerce them to smaller domains and codomains,
as the inconsistency shows. On the other hand a HOL system
with product types would have types such as function space A -> B.
Dan Christensen
Dec 10, 2021, 11:54:13 PM12/10/21
to
On Friday, December 10, 2021 at 6:29:21 PM UTC-5, Mostowski Collapse wrote:
> Dan Christensen schrieb am Freitag, 10. Dezember 2021 um 23:28:27 UTC+1:
> > On Friday, December 10, 2021 at 1:07:48 PM UTC-5, Mostowski Collapse wrote:
> > > Dan-O-Matik halucinated:
> > > > "f(x)" makes no sense if you have not specified a domain for x.
> > > I have specified a domain, it is P. What is wrong with you?
> > > The strenghening of the inconsistency is:
> > > > ∀x(Px → ∃y(Qy ∧ f(x)=y)) → (∀x∀y(f(x)=y → (Px ∧ Qy)) ↔ ∀xPx) is valid.
> > > https://www.umsu.de/trees/#~6x%28Px~5~7y%28Qy~1f%28x%29=y%29%29~5%28~6x~6y%28f%28x%29=y~5Px~1Qy%29~4~6xPx%29
> > > ∀x(Px → ∃y(Qy ∧ f(x)=y)) is "serial", Dan-O-Matiks favorite
> > > ∀x∀y(f(x)=y → (Px ∧ Qy)) is "binary relation",
> > > ∀xPx is universal class
> > >
> > > For "serial" and "binary relation" see wiki:
> > > https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
> > So, where is this "inconsistency" in DC Proof, Jan Burse? Recall, you will need a theorem in DC Proof of the form P & ~P. Have you given up already? Oh, well...
> >
> Dan Christensen schrieb am Freitag, 10. Dezember 2021 um 23:28:27 UTC+1:
> > On Friday, December 10, 2021 at 1:07:48 PM UTC-5, Mostowski Collapse wrote:
> > > Dan-O-Matik halucinated:
> > > > "f(x)" makes no sense if you have not specified a domain for x.
> > > I have specified a domain, it is P. What is wrong with you?
> > > The strenghening of the inconsistency is:
> > > > ∀x(Px → ∃y(Qy ∧ f(x)=y)) → (∀x∀y(f(x)=y → (Px ∧ Qy)) ↔ ∀xPx) is valid.
> > > https://www.umsu.de/trees/#~6x%28Px~5~7y%28Qy~1f%28x%29=y%29%29~5%28~6x~6y%28f%28x%29=y~5Px~1Qy%29~4~6xPx%29
> > > ∀x(Px → ∃y(Qy ∧ f(x)=y)) is "serial", Dan-O-Matiks favorite
> > > ∀x∀y(f(x)=y → (Px ∧ Qy)) is "binary relation",
> > > ∀xPx is universal class
> > >
> > > For "serial" and "binary relation" see wiki:
> > > https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
> > So, where is this "inconsistency" in DC Proof, Jan Burse? Recall, you will need a theorem in DC Proof of the form P & ~P. Have you given up already? Oh, well...
> >
> I nowhere wrote inconsistency "in DC Proof" in this thread.
So, you did not find an inconsistency in DC Proof. Whew!
> > Try adding these axioms:
> >
> > 1) ∃x¬Px
> > P is not the universal class.
> >
> > 2) ∀x∀y(f(x)=y → (Px ∧ Qy))
Try again.
> > f is a binary relation
> >
> > You can derive an inconsistency.
> The tree tool can do it:
>
> ∃x¬Px, ∀x∀y(f(x)=y → (Px ∧ Qy)) entails R∧¬R.
Dan Christensen
Dec 11, 2021, 12:46:40 AM12/11/21
to
1 ALL(a):[a in x => f(a) in y]
Axiom
2 t in x
Premise
3 t in x => f(t) in y
U Spec, 1
4 f(t) in y
Detach, 3, 2
5 f(t)=f(t)
Reflex, 4
6 u in y
Premise
7 u=f(t)
Premise
8 u=f(t) => u=f(t)
Conclusion, 7
9 ALL(c):[c in y => [c=f(t) => c=f(t)]]
Conclusion, 6
10 f(t) in y & f(t)=f(t)
Join, 4, 5
11 f(t) in y & f(t)=f(t) & ALL(c):[c in y => [c=f(t) => c=f(t)]]
Join, 10, 9
12 EXIST(b):[b in y & b=f(t) & ALL(c):[c in y => [c=f(t) => c=b]]]
E Gen, 11
13 ALL(a):[a in x => EXIST(b):[b in y & b=f(a) & ALL(c):[c in y => [c=f(a) => c=b]]]]
Conclusion, 2
Message has been deleted
Mostowski Collapse
Dec 11, 2021, 2:52:53 AM12/11/21
to
Its not a prof that says anything about the notion "binary relation",
which is the subject matter of this thread. You know the thread
asks the following:
"What is the truth value of f(x)=y if x is outside the domain of the function f?"
Where do you prove something about f ⊆ x x y ? You proof doesn't
end with a formula that contains f ⊆ x x y. But we can prove the
following. From:
(*) ALL(a):[a in x => f(a) in y]
We can derive f ⊆ x x y iff x=V:
(**) ALL(a):ALL(b):[f(a)=b => a in x & b in y] <=> ALL(a):[a in x]]
Good luck with finding a set such that ALL(a):[a in x], If you
combine the above with your proof of impossiblity of the
universal set, you can prove:
(***) ~ALL(a):ALL(b):[f(a)=b => a in x & b in y]
P.S.: Here is a mechanized proof of (*) -> (**), I am using E__ for _ in _:
∀a(Eax → Ef(a)y) entails ∀a∀b(f(a)=b → (Eax ∧ Eby)) ↔ ∀aEax.
https://www.umsu.de/trees/#~6a%28Eax~5Ef%28a%29y%29|=%28~6a~6b%28f%28a%29=b~5Eax~1Eby%29~4~6aEax%29
Dan Christensen halucinated:
which is the subject matter of this thread. You know the thread
asks the following:
"What is the truth value of f(x)=y if x is outside the domain of the function f?"
Where do you prove something about f ⊆ x x y ? You proof doesn't
end with a formula that contains f ⊆ x x y. But we can prove the
following. From:
(*) ALL(a):[a in x => f(a) in y]
We can derive f ⊆ x x y iff x=V:
(**) ALL(a):ALL(b):[f(a)=b => a in x & b in y] <=> ALL(a):[a in x]]
Good luck with finding a set such that ALL(a):[a in x], If you
combine the above with your proof of impossiblity of the
universal set, you can prove:
(***) ~ALL(a):ALL(b):[f(a)=b => a in x & b in y]
P.S.: Here is a mechanized proof of (*) -> (**), I am using E__ for _ in _:
∀a(Eax → Ef(a)y) entails ∀a∀b(f(a)=b → (Eax ∧ Eby)) ↔ ∀aEax.
https://www.umsu.de/trees/#~6a%28Eax~5Ef%28a%29y%29|=%28~6a~6b%28f%28a%29=b~5Eax~1Eby%29~4~6aEax%29
Dan Christensen halucinated:
Dan Christensen
Dec 11, 2021, 10:56:38 AM12/11/21
to
On Saturday, December 11, 2021 at 2:52:53 AM UTC-5, Mostowski Collapse wrote:
> Its not a prof that says anything about the notion "binary relation",
> which is the subject matter of this thread. You know the thread
> asks the following:
>
> "What is the truth value of f(x)=y if x is outside the domain of the function f?"
>
> Where do you prove something about f ⊆ x x y ?
I do better than that. I prove that for all a in x, the exists a unique value f(a) in y.
> Its not a prof that says anything about the notion "binary relation",
> which is the subject matter of this thread. You know the thread
> asks the following:
>
> "What is the truth value of f(x)=y if x is outside the domain of the function f?"
>
> Where do you prove something about f ⊆ x x y ?
ALL(a):[a in x => EXIST(b):[b in y & b=f(a) & ALL(c):[c in y => [c=f(a) => c=b]]]]
Mostowski Collapse
Dec 11, 2021, 11:03:07 AM12/11/21
to
This does not say something about:
"What is the truth value of f(x)=y if x is outside the domain of the function f?"
It only talks about inside the domain.
"What is the truth value of f(x)=y if x is outside the domain of the function f?"
Dan Christensen
Dec 11, 2021, 11:22:48 AM12/11/21
to
On Saturday, December 11, 2021 at 11:03:07 AM UTC-5, Mostowski Collapse wrote:
> Dan Christensen schrieb am Samstag, 11. Dezember 2021 um 16:56:38 UTC+1:
> > On Saturday, December 11, 2021 at 2:52:53 AM UTC-5, Mostowski Collapse wrote:
> > > Its not a prof that says anything about the notion "binary relation",
> > > which is the subject matter of this thread. You know the thread
> > > asks the following:
> > >
> > > "What is the truth value of f(x)=y if x is outside the domain of the function f?"
> > >
> > > Where do you prove something about f ⊆ x x y ?
> > I do better than that. I prove that for all a in x, the exists a unique value f(a) in y.
> > ALL(a):[a in x => EXIST(b):[b in y & b=f(a) & ALL(c):[c in y => [c=f(a) => c=b]]]]
> > However, it is also possible to construct, i.e. prove the existence of {(a,b) in xy : f(a)=b} where xy is the Cartesian product of x and y. Unfortunately for you, you cannot use it to prove f(a)=/=b for a not in x. And that's a good thing if you want to talk about the usual functions in most math textbooks, and not just some quirky, ivory-tower construct.
> This does not say something about:
> "What is the truth value of f(x)=y if x is outside the domain of the function f?"
> It only talks about inside the domain.
So, you are no longer claiming that f(x)=/=y for x not in the domain of f. Very good! As in just about every math textbook on the planet, it is simply undefined.
> Dan Christensen schrieb am Samstag, 11. Dezember 2021 um 16:56:38 UTC+1:
> > On Saturday, December 11, 2021 at 2:52:53 AM UTC-5, Mostowski Collapse wrote:
> > > Its not a prof that says anything about the notion "binary relation",
> > > which is the subject matter of this thread. You know the thread
> > > asks the following:
> > >
> > > "What is the truth value of f(x)=y if x is outside the domain of the function f?"
> > >
> > > Where do you prove something about f ⊆ x x y ?
> > I do better than that. I prove that for all a in x, the exists a unique value f(a) in y.
> > ALL(a):[a in x => EXIST(b):[b in y & b=f(a) & ALL(c):[c in y => [c=f(a) => c=b]]]]
> > However, it is also possible to construct, i.e. prove the existence of {(a,b) in xy : f(a)=b} where xy is the Cartesian product of x and y. Unfortunately for you, you cannot use it to prove f(a)=/=b for a not in x. And that's a good thing if you want to talk about the usual functions in most math textbooks, and not just some quirky, ivory-tower construct.
> This does not say something about:
> "What is the truth value of f(x)=y if x is outside the domain of the function f?"
> It only talks about inside the domain.
Mostowski Collapse
Dec 11, 2021, 11:27:27 AM12/11/21
to
For F = {(a,b) in x x y : f(a) = b } you can prove:
ALL(a):ALL(b):[(a,b) e F => a in x & b in y]
But you cannot prove F = f. But for math text book, when we talk
about a function graph, we mean F and not f.
Most COMBINATORIAL MATHEMATICS theorems work only for
function graphs, dont work for Dan-O-Matik functions.
Example:
|{ f | f : {0,..,n-1} -> {0,1} )| = 2^n
Doron Zeilberger he says he is interested in combinatorial mathematics,
as an ultrafinitist he should be down to earth, and not high alof
in an ivory tower. Maybe Dan-O-Matik lives in cloud cuckoo land.
LMAO!
ALL(a):ALL(b):[(a,b) e F => a in x & b in y]
But you cannot prove F = f. But for math text book, when we talk
about a function graph, we mean F and not f.
Most COMBINATORIAL MATHEMATICS theorems work only for
function graphs, dont work for Dan-O-Matik functions.
Example:
|{ f | f : {0,..,n-1} -> {0,1} )| = 2^n
Doron Zeilberger he says he is interested in combinatorial mathematics,
as an ultrafinitist he should be down to earth, and not high alof
in an ivory tower. Maybe Dan-O-Matik lives in cloud cuckoo land.
LMAO!
Mostowski Collapse
Dec 11, 2021, 11:55:42 AM12/11/21
to
If you want to do what you want to do Dan-O-Matik,
you need to use the notation (A, B, f) and not f : A -> B.
The notation (A, B, f) is abuse of Burbaki, it doesn't
require f ⊆ A x B. You find this explained on wikipedia:
"The set G is called the graph of f. Some authors identify it with
the function; however, in common usage, the function is generally
distinguished from its graph. In this approach, a function is
defined as an ordered triple (X, Y, G). In this notation, whether a
function is surjective (see below) depends on the choice of Y."
https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
I still stand put that you use the notation f: A -> B wrongly,
its even impossible to identify such an f with a FOL function
symbol, unless A=V and B=V. If you pass around a tripple
everything would be fine. Except that (A,B,f) is more limited
than f : A -> B, because (A,B,f) doesn't require binary relation.
So for combinatorial mathematics, you can prove:
/* Provable */
|{ f | f : {0,..,n-1} -> {0,1} }| = 2^n
But you cannot prove:
/* Not Provable */
|{ f | ({0,..,n-1},{0,1},f) Dan-O-Matik triple }| = 2^n
you need to use the notation (A, B, f) and not f : A -> B.
The notation (A, B, f) is abuse of Burbaki, it doesn't
require f ⊆ A x B. You find this explained on wikipedia:
"The set G is called the graph of f. Some authors identify it with
the function; however, in common usage, the function is generally
distinguished from its graph. In this approach, a function is
defined as an ordered triple (X, Y, G). In this notation, whether a
function is surjective (see below) depends on the choice of Y."
https://en.wikipedia.org/wiki/Function_%28mathematics%29#Relational_approach
I still stand put that you use the notation f: A -> B wrongly,
its even impossible to identify such an f with a FOL function
symbol, unless A=V and B=V. If you pass around a tripple
everything would be fine. Except that (A,B,f) is more limited
than f : A -> B, because (A,B,f) doesn't require binary relation.
So for combinatorial mathematics, you can prove:
/* Provable */
|{ f | f : {0,..,n-1} -> {0,1} }| = 2^n
But you cannot prove:
/* Not Provable */
|{ f | ({0,..,n-1},{0,1},f) Dan-O-Matik triple }| = 2^n
Mostowski Collapse
Dec 11, 2021, 12:04:40 PM12/11/21
to
Why would a Dan-O-Matik triple be abuse of Burbaki.
Because Burbaki had it this way:
1) The triple had different placement of it elements, it would
not be (A,B,f), but rather (f,A,B).
2) Burbaki had also the requirement of binary relation, he
explained this here on PDF page 61, Definition 2:
Definition 2. - On appelle correspondence enter un esemble A
et un ensemble B un triple Γ = (G, A, B) etc..
He basically requires G ⊆ A x B and G only ordered paris.
3) On PDF page 64 he then introduces f = (F, A, B).
See for yourself:
http://tomlr.free.fr/Math%E9matiques/Bourbaki/Theorie%20Des%20Ensembles.pdf
Because Burbaki had it this way:
1) The triple had different placement of it elements, it would
not be (A,B,f), but rather (f,A,B).
2) Burbaki had also the requirement of binary relation, he
explained this here on PDF page 61, Definition 2:
Definition 2. - On appelle correspondence enter un esemble A
et un ensemble B un triple Γ = (G, A, B) etc..
He basically requires G ⊆ A x B and G only ordered paris.
3) On PDF page 64 he then introduces f = (F, A, B).
See for yourself:
http://tomlr.free.fr/Math%E9matiques/Bourbaki/Theorie%20Des%20Ensembles.pdf
Dan Christensen
Dec 11, 2021, 4:44:53 PM12/11/21
to
On Saturday, December 11, 2021 at 11:27:27 AM UTC-5, Mostowski Collapse wrote:
> > > However, it is also possible to construct, i.e. prove the existence of {(a,b) in xy : f(a)=b} where xy is the Cartesian product of x and y. Unfortunately for you, you cannot use it to prove f(a)=/=b for a not in x. And that's a good thing if you want to talk about the usual functions in most math textbooks, and not just some quirky, ivory-tower construct.
> > > However, it is also possible to construct, i.e. prove the existence of {(a,b) in xy : f(a)=b} where xy is the Cartesian product of x and y. Unfortunately for you, you cannot use it to prove f(a)=/=b for a not in x. And that's a good thing if you want to talk about the usual functions in most math textbooks, and not just some quirky, ivory-tower construct.
> For F = {(a,b) in x x y : f(a) = b } you can prove:
>
> ALL(a):ALL(b):[(a,b) e F => a in x & b in y]
>
> But you cannot prove F = f.
Not important. Important is having f(x) being undefined for x not in the domain of f, as in just about every math textbook on the planet.
>
> ALL(a):ALL(b):[(a,b) e F => a in x & b in y]
>
> But you cannot prove F = f.
[snip]
Mostowski Collapse
Dec 11, 2021, 6:11:16 PM12/11/21
to
I don't say that your approach is totally wrong. At least
the classical notion f : A -> B also obeys ∀x(x e A -> f(x) e B).
But the classical notion implies more, for example you can
prove with the classical notion where A={0,...,n-1} and B={0,1} that:
/* Provable */
|{ f | f : A -> B }| = 2^n
But you cannot prove:
/* Not Provable */
|{ f | ∀x(x e A -> f(x) e B) }| = 2^n
So if you would ask a mathematics teacher, whether
∀x(x e A -> f(x) e B) is the right translation of f : A -> B,
he would say no.
You would pass an exam with this translation. Its
the same with Russels "the X is Y", wrong is wrong.
Twice wrong doesn't give right.
the classical notion f : A -> B also obeys ∀x(x e A -> f(x) e B).
But the classical notion implies more, for example you can
prove with the classical notion where A={0,...,n-1} and B={0,1} that:
/* Provable */
|{ f | f : A -> B }| = 2^n
But you cannot prove:
/* Not Provable */
So if you would ask a mathematics teacher, whether
∀x(x e A -> f(x) e B) is the right translation of f : A -> B,
he would say no.
You would pass an exam with this translation. Its
the same with Russels "the X is Y", wrong is wrong.
Twice wrong doesn't give right.
Mostowski Collapse
Dec 11, 2021, 6:13:02 PM12/11/21
to
Corr.:
You would not pass an exam with this translation.
You would not pass an exam with this translation.
Dan Christensen
Dec 11, 2021, 10:26:11 PM12/11/21
to
On Saturday, December 11, 2021 at 6:11:16 PM UTC-5, Mostowski Collapse wrote:
> So if you would ask a mathematics teacher, whether
> ∀x(x e A -> f(x) e B) is the right translation of f : A -> B,
> he would say no.
>
From no less a teacher than Terrance Tao himself:
> So if you would ask a mathematics teacher, whether
> ∀x(x e A -> f(x) e B) is the right translation of f : A -> B,
> he would say no.
>
"Definition 3.3.1 (Functions). Let X, Y be sets, and let P(x, y) be a property pertaining to an object x ∈ X and an object y ∈ Y , such that for every x ∈ X, there is exactly one y ∈ Y for which P(x, y) is true (this is sometimes known as the vertical line test). Then we define the function f : X → Y defined by P on the domain X and range Y to be the object which, given any input x ∈ X, assigns an output f(x) ∈ Y, defined to be the unique object f(x) for which P(x, f(x)) is true.
"Thus, for any x ∈ X and y ∈ Y, y = f(x) ⇐⇒ P(x, y) is true." [Note that x is restricted here to the domain X.]
Terrance Tao, "Analysis I, 3rd Ed, p49"
https://lms.umb.sk/pluginfile.php/111477/mod_page/content/5/TerenceTao_Analysis.I.Third.Edition.pdf
There is absolutely no consideration of f(x) outside of the domain of the function f. Must be frustrating as hell for you, Jan Burse. But it does rid us of Burse's Paradox.
Mostowski Collapse
Dec 12, 2021, 4:04:34 AM12/12/21
to
Your claim is baseless. How do you prove that there are 2^n many
functions that assign the values 0 and 1 to the numbers 0,..,n-1 ?
How do you prove:
/* Provable */
|{ f | f : A -> B }| = 2^n
You would not pass an exam with your crank translation.
functions that assign the values 0 and 1 to the numbers 0,..,n-1 ?
How do you prove:
/* Provable */
|{ f | f : A -> B }| = 2^n
Mostowski Collapse
Dec 12, 2021, 4:09:52 AM12/12/21
to
Also you would need to translate Terrence Taos:
pertain means belong. We can abreviate the above
by P ⊆ X x Y, and we have the wikipedia definition of
binary relation again. You are quite dick head Dan-O-Matik
with your fake news snipping technique.
Once a crank always a crank.
"P(x, y) be a property pertaining to an object x ∈ X and an object y ∈ Y"
∀x∀y(P(x, y) → x ∈ X ∧ y ∈ Y)
pertain means belong. We can abreviate the above
by P ⊆ X x Y, and we have the wikipedia definition of
binary relation again. You are quite dick head Dan-O-Matik
with your fake news snipping technique.
Once a crank always a crank.
Mostowski Collapse
Dec 12, 2021, 4:12:31 AM12/12/21
to
Your Dan-O-Matik nonsense would be correct if Terrence Taos
would have somewhere a phrase an "arbitrary property P(x, y)".
But he expressis verbis says "P(x, y) be a property pertaining to
an object x ∈ X and an object y ∈ Y". Your Dan-O-Matik nonsense
does not get right in anyway. Two times wrong doesn't give right.
would have somewhere a phrase an "arbitrary property P(x, y)".
But he expressis verbis says "P(x, y) be a property pertaining to
an object x ∈ X and an object y ∈ Y". Your Dan-O-Matik nonsense
does not get right in anyway. Two times wrong doesn't give right.
Dan Christensen
Dec 12, 2021, 10:30:23 AM12/12/21
to
On Sunday, December 12, 2021 at 4:09:52 AM UTC-5, Mostowski Collapse wrote:
> Also you would need to translate Terrence Taos:
> "P(x, y) be a property pertaining to an object x ∈ X and an object y ∈ Y"
> ∀x∀y(P(x, y) → x ∈ X ∧ y ∈ Y)
>
In DC Proof, to formally construct, i.e. prove the existence of a function f: x --> y, we proceed as follows with the required property P being introduced on line 11:
> Also you would need to translate Terrence Taos:
> "P(x, y) be a property pertaining to an object x ∈ X and an object y ∈ Y"
> ∀x∀y(P(x, y) → x ∈ X ∧ y ∈ Y)
>
Given sets x and y
1 Set(x)
Axiom
2 Set(y)
Axiom
Apply Cartesian Product Axiom for x and y
3 ALL(a1):ALL(a2):[Set(a1) & Set(a2) => EXIST(b):[Set'(b) & ALL(c1):ALL(c2):[(c1,c2) in b <=> c1 in a1 & c2 in a2]]]
Cart Prod
4 ALL(a2):[Set(x) & Set(a2) => EXIST(b):[Set'(b) & ALL(c1):ALL(c2):[(c1,c2) in b <=> c1 in x & c2 in a2]]]
U Spec, 3
5 Set(x) & Set(y) => EXIST(b):[Set'(b) & ALL(c1):ALL(c2):[(c1,c2) in b <=> c1 in x & c2 in y]]
U Spec, 4
6 Set(x) & Set(y)
Join, 1, 2
7 EXIST(b):[Set'(b) & ALL(c1):ALL(c2):[(c1,c2) in b <=> c1 in x & c2 in y]]
Detach, 5, 6
xy is the Cartesian product of x and y
E Spec, 7
9 Set'(xy)
Split, 8
10 ALL(c1):ALL(c2):[(c1,c2) in xy <=> c1 in x & c2 in y]
Split, 8
Apply Subset Axiom to introduce binary predicate P and the set of ordered pairs f that will be shown to be a function.
11 EXIST(a):[Set'(a) & ALL(b):ALL(c):[(b,c) in a <=> (b,c) in xy & P(b,c)]]
Subset, 9
12 Set'(f) & ALL(b):ALL(c):[(b,c) in f <=> (b,c) in xy & P(b,c)]
E Spec, 11
13 Set'(f)
Split, 12
14 ALL(b):ALL(c):[(b,c) in f <=> (b,c) in xy & P(b,c)]
Split, 12
Prove: f is a function such that f: x --> y
Apply Function Axiom (for 1 variable) to establish sufficient conditions
15 ALL(f):ALL(dom):ALL(cod):[Set'(f) & Set(dom) & Set(cod)
=> [ALL(a1):ALL(b):[(a1,b) in f => a1 in dom & b in cod]
& ALL(a1):[a1 in dom => EXIST(b):[b in cod & (a1,b) in f]]
& ALL(a1):ALL(b1):ALL(b2):[a1 in dom & b1 in cod & b2 in cod
=> [(a1,b1) in f & (a1,b2) in f => b1=b2]]
=> ALL(a1):ALL(b):[a1 in dom & b in cod => [f(a1)=b <=> (a1,b) in f]]]]
Function
16 ALL(dom):ALL(cod):[Set'(f) & Set(dom) & Set(cod)
=> [ALL(a1):ALL(b):[(a1,b) in f => a1 in dom & b in cod]
& ALL(a1):[a1 in dom => EXIST(b):[b in cod & (a1,b) in f]]
& ALL(a1):ALL(b1):ALL(b2):[a1 in dom & b1 in cod & b2 in cod
=> [(a1,b1) in f & (a1,b2) in f => b1=b2]]
=> ALL(a1):ALL(b):[a1 in dom & b in cod => [f(a1)=b <=> (a1,b) in f]]]]
U Spec, 15
17 ALL(cod):[Set'(f) & Set(x) & Set(cod)
=> [ALL(a1):ALL(b):[(a1,b) in f => a1 in x & b in cod]
& ALL(a1):[a1 in x => EXIST(b):[b in cod & (a1,b) in f]]
& ALL(a1):ALL(b1):ALL(b2):[a1 in x & b1 in cod & b2 in cod
=> [(a1,b1) in f & (a1,b2) in f => b1=b2]]
=> ALL(a1):ALL(b):[a1 in x & b in cod => [f(a1)=b <=> (a1,b) in f]]]]
U Spec, 16
18 Set'(f) & Set(x) & Set(y)
=> [ALL(a1):ALL(b):[(a1,b) in f => a1 in x & b in y]
& ALL(a1):[a1 in x => EXIST(b):[b in y & (a1,b) in f]]
& ALL(a1):ALL(b1):ALL(b2):[a1 in x & b1 in y & b2 in y
=> [(a1,b1) in f & (a1,b2) in f => b1=b2]]
=> ALL(a1):ALL(b):[a1 in x & b in y => [f(a1)=b <=> (a1,b) in f]]]
U Spec, 17
19 Set'(f) & Set(x)
Join, 13, 1
20 Set'(f) & Set(x) & Set(y)
Join, 19, 2
We have 3 preconditions here for the functionality of f:
21 ALL(a1):ALL(b):[(a1,b) in f => a1 in x & b in y]
& ALL(a1):[a1 in x => EXIST(b):[b in y & (a1,b) in f]]
& ALL(a1):ALL(b1):ALL(b2):[a1 in x & b1 in y & b2 in y
=> [(a1,b1) in f & (a1,b2) in f => b1=b2]]
=> ALL(a1):ALL(b):[a1 in x & b in y => [f(a1)=b <=> (a1,b) in f]]
Detach, 18, 20
Mostowski Collapse
Dec 12, 2021, 10:49:15 AM12/12/21
to
But you cannot prove:
/* Provable */
length 10 and with element values from 0 and 1?
There are only 2^10 = 1024 such arrays. How
do you want to do this with a FOL function symbol
unless you fix for example:
(*) ALL(a1):[~a1 in {0,..,n-1} -> f(x)=⊥]
But the above (*) doesn't follow from the nonsense
you proved (picked from the Terrence Tao site):
(**) ALL(a1):ALL(b):[a1 in {0,..,n-1} & b in {0,1} => [f(a1)=b <=> (a1,b) in f]]
What you proved again talks only inside the domain
and co-domain of the function. You didn't prove or
show something useful as per the topic of this thread here:
"What is the truth value of f(x)=y if x is outside the domain of the function f?"
/* Provable */
|{ f | f : {0,..,n-1} -> {0,1} }| = 2^n
In JavaScript how many arrays are there with
length 10 and with element values from 0 and 1?
There are only 2^10 = 1024 such arrays. How
do you want to do this with a FOL function symbol
unless you fix for example:
(*) ALL(a1):[~a1 in {0,..,n-1} -> f(x)=⊥]
But the above (*) doesn't follow from the nonsense
you proved (picked from the Terrence Tao site):
(**) ALL(a1):ALL(b):[a1 in {0,..,n-1} & b in {0,1} => [f(a1)=b <=> (a1,b) in f]]
What you proved again talks only inside the domain
and co-domain of the function. You didn't prove or
show something useful as per the topic of this thread here:
"What is the truth value of f(x)=y if x is outside the domain of the function f?"
Dan Christensen
Dec 12, 2021, 10:51:32 AM12/12/21
to
ALL(a):ALL(b):[a in x & b in y => [b=f(a) <=> P(a,b)]]
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