How the hell can q -> q??????
Stonelock
mean anything semantically?
A cause can never be its consequence!!!!!!
Stonelock
Mike Oliver
Material implication has precious little to do with causality.
The semantic meaning of "q -> q" is not "q causes q" but
rather "q is not true, or q is true", where the "or" is
inclusive (even though in this particular case it doesn't matter).
There are other semantic interpretations you might try. An
intuitionist, for example, would read "q -> q" as "there
is an effective method that turns any proof of q into
a proof of q".
I don't know of any systematic treatment that inteprets -> as
causal implication, though I'm sure that someone must have
come up with one. Actually that's probably an underestimate;
likely there are many different competing systems with such
a motivation over which their proponents have endless and
incomprehensible arguments.
Thad Coons
"Mike Oliver" <oli...@math.ucla.edu> wrote in message
news:3DD43882...@math.ucla.edu...
You could also try P ->P as " P is at least as true as P"
Thad Coons
G. Frege
(Stonelock) wrote:
>
> How the hell can you prove syntactically that q -> q when it doesn't
> mean anything semantically?
>
Who (the hell) said that it doesn't mean _anything_ semantically?!
Of course you have to plug in something for "q" first. Say, "it is
day". Then we get: "If it is day, it is day". Obviously a _true_
sentence, no?
Actually it's a tautology: with other words: for any P we have: If P,
than P. It's a "basic" (logical) truth.
>
> A cause can never be its consequence!!!
>
You obviously never heard the term "causa sui", did you? :-)
Ok, let's assume that this is true, what you say. Now how is this
related with "p->p"? It is NOT. "p->p" is just a SENTENCE or
PROPOSITION like any other. The "->" is somehow MISLEADING here.
If I write it THIS WAY "~p v p", you surely will have less problems:
Either not-p or p.
A basic "law" (of classical logic).
F.
G. Frege
wrote:
>
> You could also try P -> P as "P is at least as true as P"
>
Quite an interesting interpretation.
You mean: "P -> Q" ... "Q is at least as true as P", right?
Under this interpretation we would have the usual truth-table for
classical two-valued logic
P Q P -> Q
T T T
T F F
F T T
F F T
...but the definition would certainly be much more "general"...
(interesting, really).
If we had "grades" of truth(ness). "P->Q" would ensure at least the
"propagation" of (at least that) truth-value that P has.
Actually your interpretation seems to be in good agreement with the
usual "usage" of "If so and so, then so and so" in natural language.
F.
Stonelock
>
> > How the hell can you prove syntaxically that q -> q when it doesnt
> > mean anything semantically?
> > A cause can never be its consequence!!!!!!
>
> Material implication has precious little to do with causality.
> The semantic meaning of "q -> q" is not "q causes q" but
> rather "q is not true, or q is true", where the "or" is
> inclusive (even though in this particular case it doesn't matter).
Thats the reason of my intervention. Its can't imagine that it would
be accepted as a valid result in any system. I never agreed with it.
Even if it only involves machanical syntaxic manipulations. I don't
think such things should be allowed.
> There are other semantic interpretations you might try. An
> intuitionist, for example, would read "q -> q" as "there
> is an effective method that turns any proof of q into
> a proof of q".
Hehe, no, i don't agree with the notion period :).
> I don't know of any systematic treatment that inteprets -> as
> causal implication, though I'm sure that someone must have
> come up with one. Actually that's probably an underestimate;
> likely there are many different competing systems with such
> a motivation over which their proponents have endless and
> incomprehensible arguments.
I dont think any operator pretending implication should be anything
else than causal implication :) hehe.
Stonelock
tc...@lsa.umich.edu
Stonelock <crypto_s...@hotmail.com> wrote:
>Thats the reason of my intervention. Its can't imagine that it would
>be accepted as a valid result in any system. I never agreed with it.
>Even if it only involves machanical syntaxic manipulations. I don't
>think such things should be allowed.
In other words, if "q -> q" makes any sense, then you're a stuffed monkey.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
Thad Coons
"G. Frege" <in...@simple-line.de> wrote in message
news:9dr8tu0p4f55lb51q...@4ax.com...
> On Thu, 14 Nov 2002 18:28:47 -0600, "Thad Coons" <toc...@citlink.net>
> wrote:
>
> >
> > You could also try P -> P as "P is at least as true as P"
> >
> Quite an interesting interpretation.
>
> You mean: "P -> Q" ... "Q is at least as true as P", right?
Exactly
> Under this interpretation we would have the usual truth-table for
> classical two-valued logic
>
> P Q P -> Q
>
> T T T
> T F F
> F T T
> F F T
>
> ...but the definition would certainly be much more "general"...
> (interesting, really).
You never know what someone else is going to find interesting.
When I began working on 3VL, in ignorance of what had already been done, I
couldn't even prove P->P, since my definitions didn't work for the middle
value. When I started doing some reasearch and found everything I had done
had been done before, (notably, Lukasieiwicz did better, but not quite good
enough).
When I finally noticed this possible interpretation, I was enormously
pleased with how well it worked. Everything I had been trying to do in 3VL
fell into place.
> If we had "grades" of truth(ness). "P->Q" would ensure at least the
> "propagation" of (at least that) truth-value that P has.
>
> Actually your interpretation seems to be in good agreement with the
> usual "usage" of "If so and so, then so and so" in natural language.
I think it's also in decent agreement with how logic is commonly used in
mathematics, for instance. Logic isn't particularly good at coming up with
theorems, but it's excellent as a test for whether something is "at least as
true" as the axioms. Given that we have claimed our axioms to be true, we
can be reliably assured that the theorems are at least that true.
Thad Coons
Stonelock
> wrote:
>
> >
> > You could also try P -> P as "P is at least as true as P"
> >
> Quite an interesting interpretation.
>
> You mean: "P -> Q" ... "Q is at least as true as P", right?
>
> Under this interpretation we would have the usual truth-table for
> classical two-valued logic
>
> P Q P -> Q
>
> T T T
> T F F
> F T T
> F F T
>
> ...but the definition would certainly be much more "general"...
> (interesting, really).
>
> If we had "grades" of truth(ness).
Yeahs, different semantic levels :).
"P->Q" would ensure at least the
> "propagation" of (at least that) truth-value that P has.
With an important restriction!!!. P could not propagate its value to
itself. P would not be part of the set where the truth value would be
propagated.
>
> Actually your interpretation seems to be in good agreement with the
> usual "usage" of "If so and so, then so and so" in natural language.
>
> F.
I've studied propositionnal logic; what i'm saying is that i dont
agree that both arguments should be treated as being on the same
semantic level.
Stonelock
Stonelock
> In article <64f2b1e9.02111...@posting.google.com>,
> Stonelock <crypto_s...@hotmail.com> wrote:
> >Thats the reason of my intervention. Its can't imagine that it would
> >be accepted as a valid result in any system. I never agreed with it.
> >Even if it only involves machanical syntaxic manipulations. I don't
> >think such things should be allowed.
>
> In other words, if "q -> q" makes any sense, then you're a stuffed monkey.
Haha, yeah, either using causal implication or propositionnal
implication ;).
The problem is that propositionnal implication allows for any 2
non-linked propositions to be evaluated on the same level while in
fact, i think the truth value of the second argument is dependant on
that of the first.
Stonelock
G. Frege
wrote:
> >
> > Under this interpretation we would [ first ] have the usual truth-table for
> > classical two-valued logic
>>
> > [ but second ] If we had "grades" of truth(ness). "P->Q" would ensure at least the
> > "propagation" of (at least that) truth-value that P has.
> >
>
> [...] When I finally noticed this possible interpretation, I was enormously
> pleased with how well it worked. Everything I had been trying to do in 3VL
> fell into place.
>
I see.
> >
> > Actually your interpretation seems to be in good agreement with the
> > usual "usage" [ or even semantics ] of "If so and so, then so and so"
> > in natural language.
> >
> I think it's also in decent agreement with how logic is commonly used in
> mathematics, for instance. Logic isn't particularly good at coming up with
> theorems, but it's excellent as a test for whether something is "at least as
> true" as the axioms. Given that we have claimed our axioms to be true, we
> can be reliably assured that the theorems are at least that true.
>
Yes, sure. But that's exactly what the standard "implication" does
too.
I had _something different_ in mind here... But don't worry, be h...
F.
G. Frege
(Stonelock) wrote:
> >
> > If we had "grades" of truth(ness).
> >
> Yeah, different semantic levels :).
>
No. That was not _meant_ here.
> >
> > "P -> Q" would ensure at least the "propagation"
> > of (at least) that truth-value that P has.
> >
> With an important restriction! P could not propagate its value to
> itself.
>
*Sigh* ... Know what _metaphorical speech_ means?
Actually P (and/or "->") does NOTHING. But a formula of the FORM
"X -> Y" *expresses* that "Y" is true IF "X" is true. Now if "X" and
"Y" are replaced by the same propositional letter (or sentence
letter), say "P", this is OBVIOUSLY the case (if no self reference is
involved).
What you have in mind here is a _cause relation_ sign, imho. But well,
"->" IS NOT such a symbol.
But we could (ad hoc) INVENT
->>
here for this purpose. And, yes, THEN we could question the
expression:
P ->> P.
On the other hand SOME philosophers maintain(ed) that
god ->> god
holds.
F.
Mike Oliver
> Mike Oliver <oli...@math.ucla.edu> wrote in message news:<3DD43882...@math.ucla.edu>...
>> The semantic meaning of "q -> q" is not "q causes q" but
>> rather "q is not true, or q is true", where the "or" is
>> inclusive (even though in this particular case it doesn't matter).
>
> Thats the reason of my intervention. Its can't imagine that it would
> be accepted as a valid result in any system. I never agreed with it.
> Even if it only involves machanical syntaxic manipulations. I don't
> think such things should be allowed.
The word is "syntactic", not "syntaxic". As for the rest of
it -- who cares whether you "agree" with it? We all know
what it means, and how it behaves, and we find it useful.
That's enough for us. You don't like it? You don't have to.
tc...@lsa.umich.edu
Stonelock <crypto_s...@hotmail.com> wrote:
>tc...@lsa.umich.edu wrote in message news:<ar2ik2$i4f$2...@galois.mit.edu>...
>
>Haha, yeah, either using causal implication or propositionnal
>implication ;).
So if someone on this newsgroup manages to make sense of "q -> q," that will
*cause* you to become a stuffed monkey? *That* I'd like to see...
Dan Arkoff
I will assume that you are using the "->" symbol to represent
"material implication" in classical First Order Logic (FOL) (sometimes
a horseshoe symbol is used instead of "->" to represent material
implication). In FOL (q -> q) is equivalent to (~q v q), which is
semantically interpreted as (NOT-q OR q), with OR being inclusive
(meaning "and/or"). So material implication _does_ have a meaningful
semantic interpretation.
However, the material implication (q -> q) is also often interpreted
as "IF q THEN q", but material implication clearly doesn't have the
same meaning that conditionals (IF...THEN statements) typically have
in ordinary English usage. Thus, I agree that there is something
wrong with the interpretation of (q -> q) as (IF q THEN q). But you
are mistaken in assuming that a conditional must represent cause and
effect. The meaning of conditionals in ordinary English is that there
is a connection in meaning between the antecedent (the IF clause) and
the consequent (the THEN clause) such that whenever the antecedent is
true, the consequent must also be true. But this does not require
that the antecedent _causes_ the consequent. For example, the
statement "If Fido is a dog, then Fido is a mammal" is clearly true,
because there is a connection in meaning between being a dog and being
a mammal, such that if something is a dog then it must be a mammal
(because the set of dogs is a subset of the set of mammals). But it
wouldn't really be accurate to say that Fido being a dog _causes_ Fido
to be a mammal.
In contrast, the statement "If Fido is a dog, then Fido is a reptile"
would be considered to be false in ordinary English usage, although it
_could_ be considered to be true if it were interpreted as a material
conditional (for example, if Fido is actually a fish, then the
antecedent in the above statement would be false, so under the
truth-value definitions for material implication, the statement would
be considered to be true).
The concept of cause and effect is more in the realm of science than
in the realm of deductive logic, and the concept is governed more by
_inductive_ logic (in which the conclusions do not necessarily follow
from the premises, but rather are only considered to be likely to be
true given the premises) than by _deductive_ logic.
Dan Arkoff
Stonelock
> On 15 Nov 2002 11:05:04 -0800, crypto_s...@hotmail.com
> (Stonelock) wrote:
>
> > >
> > > If we had "grades" of truth(ness).
> > >
> > Yeah, different semantic levels :).
> >
> No. That was not _meant_ here.
What did you mean then?
> > >
> > > "P -> Q" would ensure at least the "propagation"
> > > of (at least) that truth-value that P has.
> > >
> > With an important restriction! P could not propagate its value to
> > itself.
> >
> *Sigh* ... Know what _metaphorical speech_ means?
If you speak clearly then i can understand you.
> Actually P (and/or "->") does NOTHING. But a formula of the FORM
> "X -> Y" *expresses* that "Y" is true IF "X" is true.
Wrong! You do not understand boolean implication correctly. Both X and
Y are independant in boolean implication and are treated on the same
level. The value of Y is not determined by the value of X. That is
what i actually don't agree with! What i'm saying is that the boolean
implication operator represents nothing and doesn't come close to
causal implication. It is nonetheless used as such.
Now if "X" and
> "Y" are replaced by the same propositional letter (or sentence
> letter), say "P", this is OBVIOUSLY the case (if no self reference is
> involved).
>
> What you have in mind here is a _cause relation_ sign, imho. But well,
> "->" IS NOT such a symbol.
>
> But we could (ad hoc) INVENT
>
> ->>
>
> here for this purpose. And, yes, THEN we could question the
> expression:
>
> P ->> P.
This would also be absurd because P can not cause P.
> On the other hand SOME philosophers maintain(ed) that
>
> god ->> god
>
god ->> god is founded on irrationnal premisses; it is junk. God is a
human concept, not an objective reality.
Stonelock.
Thad Coons
"Stonelock" <crypto_s...@hotmail.com> wrote in message
news:64f2b1e9.02111...@posting.google.com...
> G. Frege <in...@simple-line.de> wrote in message
news:<9dr8tu0p4f55lb51q...@4ax.com>...
> > On Thu, 14 Nov 2002 18:28:47 -0600, "Thad Coons" <toc...@citlink.net>
> > wrote:
> >
> > >
> > > You could also try P -> P as "P is at least as true as P"
> > >
> > Quite an interesting interpretation.
> >
> > You mean: "P -> Q" ... "Q is at least as true as P", right?
> >
> > Under this interpretation we would have the usual truth-table for
> > classical two-valued logic
> >
> > P Q P -> Q
> >
> > T T T
> > T F F
> > F T T
> > F F T
> >
> > ...but the definition would certainly be much more "general"...
> > (interesting, really).
> >
> > If we had "grades" of truth(ness).
>
> Yeahs, different semantic levels :).
>
> "P->Q" would ensure at least the
> > "propagation" of (at least that) truth-value that P has.
>
> With an important restriction!!!. P could not propagate its value to
> itself. P would not be part of the set where the truth value would be
> propagated.
I don't know about semantic levels. "P -> Q" meaning "Q is at least as true
as P" works for P -> P. In some logics, it's defined or declared so as to
include that specific case, amont others. There are various logics in which
P ->P doesn't hold, but they typically don't have nearly the utility of
classical logic.
Thad Coons
Thad Coons
"Dan Arkoff" <DAr...@aol.com> wrote in message
news:63dc2a8a.02111...@posting.google.com...
"If Fido is a dog, then Fido is a reptile" says nothing about what
happens if Fido is not a dog. The formal truth of the material conditional
when Fido is a fish is unimportant. You can conclude nothing from it that
you didn't know already. I don't think that's a particularly strong
objection.
I take a more signicant objection to equating "if...then" with the
material conditional to come from the fact that natural language isn't
two-valued. Open, ambiguous, and contradictory statements are hard to
evaluate as "true" or "false". Numerous "if..then" statements fall in the
same category. Attempts to formalize the intuitive rules for reasoning with
doubtful and uncertain statements have met with only limited success.
Thad Coons
G. Frege
(Stonelock) wrote:
>
> The [truth] value of Y is not determined by the [truth] value of X.
>
Nonsense.
IF "X -> Y" is considered true, AND X is considered true THEN Y _must
be_ also considered true.
That's so, because "X -> Y" actually _expresses_ that "Y" is true IF
"X" is true. Simple as that.
Look, Stonehead, if you have "P->Q" and "P" you can validly derive
(deduce, conclude) "Q". [ This "principle" of valid logical reasoning
is called _modus ponens_, BTW. ]
>
> That is what i actually don't agree with!
>
Nobody is interested in what you "agree with", or not.
>
> What i'm saying is that the boolean implication operator
> represents nothing and doesn't come close to
> causal implication.
>
The first part is of this sentence is NONSENSE, the second part is
obviously true.
Of course implication represents SOMETHING. A certain _logical_
relation concerning the sentences "connected" by it.
It "P->Q" _expresses_ that the following IS NOT the case:
P true and
Q false.
Actually a logical ARGUMENT of the form "A |- B" is valid if and only
if "A -> B" is always true [...].
On the other hand, Stonehead, WHY - the hell - do you think the "->"
is called "IMPLICATION" (or _material_ implication sometimes) and NOT
"_causal_ implication"???!!!
> >
> > But we could (ad hoc) INVENT
> >
> > ->>
> >
> > here for
expressing a casual relation.
> >
> > And, yes, THEN we could question the
> > expression:
> >
> > P ->> P.
>
> This would [...] be absurd because P can not cause P.
>
Yes, this is what you CLAIM. (And _probably_ it is a reasonable
position.)
> >
> > On the other hand SOME philosophers maintain(ed) that
> >
> > god ->> god
> >
>
> god ->> god is founded on irrational premisses; it is junk. God is a
> human concept, not an objective reality.
>
That's an interesting position.
Actually this (your) statement triggered an idea [ in my mind ] :-)
BASED on the (concept of and) notation of "->>" we could actually
DEFINE "god" in the following way:
(ix) x ->> x.
God is _the_ x such that x ->> x holds. With other words: God is (the)
causa sui.
Now we could EXPRESS the claim that god actually exists in the
following way:
(Ex) x ->> x (*)
And the claim that there is exactly one god:
(Ex)(x ->> x & (Ay)(y ->> y -> x = y)). (**)
Note that there is no "irrational premisses, junk" etc. is involved
here, so far.
Now IF we could show (**) then
G = (ix) x ->> x.
actually would define ("name") a certain entity.
So in any case you could take
x ->> x
as a DESCRIPTION in Russellian sense representing the term "God".
With other words, _then_ you could meaningful express:
"God does not exist" (a)
or
"God is great" (b)
etc.
This sentences would _then_ be translated:
(a) ~(Ex)(x ->> x & (Ay)(y ->> y -> x = y))
(b) (Ex)(x ->> x & (Ay)(y ->> y -> x = y) & x is great)
Note that NO "irrational premisses, junk" etc is involved here;
actually the (questionable) term "God" has been removed from this
_claims_.
Now IF we could actually show that
(Ex)(x ->> x & (Ay)(y ->> y -> x = y))
holds; and we define
god = (ix) x ->> x
THEN we could _meaningful_ express:
god ->> god.
Moreover IN THIS CASE it would be true.
:-)
F.
G. Frege
wrote:
> > >
> > > "P->Q" would ensure at least the "propagation"
> > > of (at least) that truth-value that P has.
> > >
> > With an important restriction: P could not propagate its value to
> > itself. (...)
> >
Let's consider classical (two valued) logic.
Next time this troll will claim that
A <-> B
doesn't express that A and B have (always) the same truth value; and
hence
A <-> A
cannot be true, etc... *sigh*
But never mind, I find his reservations inspiring non the less. :-)
>
> [...] "P -> Q" meaning "Q is at least as true as P" works for P -> P.
>
Sure. [ Some even call this tautology "the law of identity". ]
F.
G. Frege
>
> In contrast, the statement "If Fido is a dog, then Fido is a reptile"
> would be considered to be false in ordinary English usage
>
[ IF Fido actually is not a dog ]
> although it _could_ be considered to be true if it were interpreted
> as a material conditional (for example, if Fido is actually a fish, then the
> antecedent in the above statement would be false, so under the
> truth-value definitions for material implication, the statement would
> be considered to be true).
>
Well, actually I have no problem to consider such a sentence to be
true.
SOMETIMES such sentences (with false antecedent, and still claimed to
hold) _are_ actually used in normal language!
For example, one could say:
If Stonelock is not a troll, I will eat my hat!
And obviously I want not express here that I am going to eat my hat.
;-)
Well, simpler formulation:
I will eat my hat, if Stonelock is really serious,
And actually I really want to express something with this claim,
namely that Stonelock (imho) is NOT really serious. (Since I clearly
have no desire to eat my hat.)
F.
G. Frege
wrote:
>
> Now we could EXPRESS the claim that god actually exists in the
> following way:
>
> (Ex)(x ->> x)
>
Now your claim is:
~(Ex)(x ->> x) ,
[ ...there is no consequence that is its own cause.*) ]
or
(x)~(x ->> x).
[ "A cause can never be its consequence." (Stonelock) ]
Now this actually would imply that there _is no_ god (defined as causa
sui).
Still we have
a = a ,
if "a" is a (denoting) name/term and
P -> P ,
if "P" is a proposition letter.
F.
*) Terminology:
If we have A ->> B we say: A is the cause of B, and B is the
consequence of A.
If we have A ->> x for some x we say: A is a cause. And if we have
y -> B for some y we say: B is a consequence.
Stonelock
> In article <64f2b1e9.02111...@posting.google.com>,
> Stonelock <crypto_s...@hotmail.com> wrote:
> >tc...@lsa.umich.edu wrote in message news:<ar2ik2$i4f$2...@galois.mit.edu>...
> >> In other words, if "q -> q" makes any sense, then you're a stuffed monkey.
> >
> >Haha, yeah, either using causal implication or propositionnal
> >implication ;).
>
> So if someone on this newsgroup manages to make sense of "q -> q," that will
> *cause* you to become a stuffed monkey? *That* I'd like to see...
So would I. If you can prove it, I'll turn into a stuffed monkey right
away. Of course that won't happen so....:).
Stonelock
Stonelock
> crypto_s...@hotmail.com (Stonelock) wrote in message news:<64f2b1e9.02111...@posting.google.com>...
> > How the hell can you prove syntaxically that q -> q when it doesnt
> > mean anything semantically?
> > A cause can never be its consequence!!!!!!
> >
> > Stonelock
>
> I will assume that you are using the "->" symbol to represent
> "material implication" in classical First Order Logic (FOL) (sometimes
> a horseshoe symbol is used instead of "->" to represent material
> implication). In FOL (q -> q) is equivalent to (~q v q), which is
> semantically interpreted as (NOT-q OR q), with OR being inclusive
> (meaning "and/or"). So material implication _does_ have a meaningful
> semantic interpretation.
You assumed right also :). But i still do not agree that material
implication does have semantic meaning.
> However, the material implication (q -> q) is also often interpreted
> as "IF q THEN q", but material implication clearly doesn't have the
> same meaning that conditionals (IF...THEN statements) typically have
> in ordinary English usage. Thus, I agree that there is something
> wrong with the interpretation of (q -> q) as (IF q THEN q). But you
> are mistaken in assuming that a conditional must represent cause and
> effect. The meaning of conditionals in ordinary English is that there
> is a connection in meaning between the antecedent (the IF clause) and
> the consequent (the THEN clause) such that whenever the antecedent is
> true, the consequent must also be true. But this does not require
> that the antecedent _causes_ the consequent. For example, the
> statement "If Fido is a dog, then Fido is a mammal" is clearly true,
> because there is a connection in meaning between being a dog and being
> a mammal, such that if something is a dog then it must be a mammal
> (because the set of dogs is a subset of the set of mammals). But it
> wouldn't really be accurate to say that Fido being a dog _causes_ Fido
> to be a mammal.
This is the best intervention i've see in the sci forum, and i've been
around for a lil while. Thank you for this perl of clarity. Actually,
this is why i disagree with material implication. It doesn't allow for
correct evaluation, of non connected and connected sets; as a result,
different semantic levels end up being interpreted as if they were on
the same level. In the case of "If "Fido is a dog, then Fido is a
mammal", there is a clear relationship because the subset of dogs is
contained in the set of mammals. But what happens when you compare 2
sets that are not related to one another? You get junk even if both
truth values can be objectively verified to yield true. Also, I'd like
to add that causal implication goes both ways. Either it allows to
"discover" a superset in which the set is part of, or it allows to
"discover" subsets of a set. Either you go up, or you go down :). I
agree that it would not be accurate to say that A causes B; it would
be more accurate to say that A means B in the case where B can be
derived from A, also A is englobed by B in the case where a is a
subset of a larger set. (i.e. As in the case of the subset of dogs
contained in the set of mammals).
> In contrast, the statement "If Fido is a dog, then Fido is a reptile"
> would be considered to be false in ordinary English usage, although it
> _could_ be considered to be true if it were interpreted as a material
> conditional (for example, if Fido is actually a fish, then the
> antecedent in the above statement would be false, so under the
> truth-value definitions for material implication, the statement would
> be considered to be true).
Yeahs, I find this sad because different frames of reference that are
not related are evaluated on the same level and I find that only junk
can emerge from such a process.
> The concept of cause and effect is more in the realm of science than
> in the realm of deductive logic, and the concept is governed more by
> _inductive_ logic (in which the conclusions do not necessarily follow
> from the premises, but rather are only considered to be likely to be
> true given the premises) than by _deductive_ logic.
I think they should be seen as one and the same :). A single operator
that either allows to find supersets, or to find subsets. The only
thing that needs to be done for such a process to work is that the
sets involved in the evaluation need to be correctly defined before
the problem is stated.
Stonelock
G. Frege
(Stonelock) wrote:
> >
> > So if someone on this newsgroup manages to make sense of "q -> q" that will
> > *cause* you to become a stuffed monkey? *That* I'd like to see...
> >
> So would I. If you can prove it, I'll turn into a stuffed monkey right
> away.
>
Fine then. The following actually the "semantical meaning" of "->":
The expression "A -> Q" is false if "A" is true and "B" is
false, it is true otherwise.
So you are a stuffed monkey now, right? :-)
F.
P.S.
Of course, next you will claim "But with 'meaning' I meant...", etc.
Hell!!! Fuck up, troll!
G. Frege
wrote:
Typo:
>
> The following actually >>IS<< the "semantical meaning" of "->"...
>
At least for most logicians (if "->" is meant to denote the usual
_implication_ of classical logic).
F.
George Dance
> Mike Oliver <oli...@math.ucla.edu> wrote in message news:<3DD43882...@math.ucla.edu>...
> > Stonelock wrote:
> >
> > > How the hell can you prove syntaxically that q -> q when it doesnt
> > > mean anything semantically?
> > > A cause can never be its consequence!!!!!!
> >
> > Material implication has precious little to do with causality.
> > The semantic meaning of "q -> q" is not "q causes q" but
> > rather "q is not true, or q is true", where the "or" is
> > inclusive (even though in this particular case it doesn't matter).
>
> Thats the reason of my intervention. Its can't imagine that it would
> be accepted as a valid result in any system. I never agreed with it.
> Even if it only involves machanical syntaxic manipulations. I don't
> think such things should be allowed.
There's nothing wrong with your coming up with your own system of
logic dealing with physical causation. (You could start by checking
out one of the existing logics, such as dynamic logic, that does deal
with physical change.)
If you wanted, you could even use all of the notation of PC in your
logic, as long as you defined your own language prcisely each time to
prevent confusion. If you did that, I doubt that anyone would be
objecting.
What's causing objections, I think, is your insistence that everyone
else is using PC incorrectly, because you don't accept what they mean
by the terms they use; that PC should be used for what you want it to
be use, and should not be used for what everyone else who uses PC uses
it for.
> > There are other semantic interpretations you might try. An
> > intuitionist, for example, would read "q -> q" as "there
> > is an effective method that turns any proof of q into
> > a proof of q".
>
> Hehe, no, i don't agree with the notion period :).
Do you understand what the notation means? My understanding is that
PC is the study of reasoning; the process by which people figure out,
given what little they know is true, what else is true. That notion
(which I hope is clear enough) is exactly the one captured by the PC
notion of implication.
> > I don't know of any systematic treatment that inteprets -> as
> > causal implication, though I'm sure that someone must have
> > come up with one. Actually that's probably an underestimate;
> > likely there are many different competing systems with such
> > a motivation over which their proponents have endless and
> > incomprehensible arguments.
>
> I dont think any operator pretending implication should be anything
> else than causal implication
Then you misunderstand what implication means in PC. 'A implies B'
means that if A is true, then B is also true; so if you know A, you
can also validly conclude B; and you know B, by the same evidence that
you know A.
To take the first example off the top of my head: I smell a bad smell
in my refrigerator, so I conclude that some food has gone bad in it.
I know the smell is there, so I know the bad food is there:
1. (bad smell)
2. ((bad smell)->(bad food))
------
3. (bad food)
That knowledge that there is bad food in my fridge gives me a reasoned
justification for looking for it and throwing it out.
Now you come along and tell me that that's all wrong; either I cannot
make such an inference at all, or, if I can, I cannot do so logically:
since (bad food) is the cause of (bad smell), and I am allowed to use
logic to reason only from sufficient causes to effects, and not from
effects to necessary causes.
Just why should I accept that? I know that inferences like this are
sound; why should I believe that they are not logical, just because
you say they aren't?
> ;) hehe.
If I discovered a bad smell in your fridge, and told you there was
some bad food in there; and your only response was "hehe"; why would
that change my mind?
G. Frege
wrote:
>
> 1. (bad smell)
> 2. ((bad smell)->(bad food))
> ------
> 3. (bad food)
>
There's an interesting "reverse" relation between "causation" and
implication.
If the bad food actually _causes_ the bad smell, denoted by:
(bad smell) <<- (bad food)
then the PC sentence/proposition
(bad smell) -> (bad food)
holds.
In general:
if A <<- B then A -> B. (*)
Since B can not be the cause of A and (at the same "time") A be true,
but B false. Note that <<- is not an extensional truth-function (with
other words: the truth-value of A <<- B is not determined solely by
the _truth-values_ of A and B. )
Now it's in fact reasonable to assume that bad smell in a refrigerator
is actually _caused_ by bad food (well, probably there are some other
_possible_ causes; but they usually can be neglected). Hence it's
quite reasonable to assume that
(bad smell)->(bad food))
actually holds.
So your conclusion
1. (bad smell)
2. ((bad smell)->(bad food))
------
3. (bad food)
actually makes sense here.
F.
G. Frege
wrote:
>
> What's causing objections, I think, is your insistence that [ whatever ]
>
What this idiot doesn't want to understand, is the simple fact that
the *symbol* "->" in "A -> B" and the translation "If A then B" can be
quite misleading.
The usually MEANING of "A -> B" actually is (the same as):
~A v B [ or: ~(A & ~B) ]
(That's the way Russell and Whitehead defined "->" in the PM.)
Hence "q -> q" has the same meaning as
~q v q.
So his "question" actually is:
"How the hell can ~q v q?????"
Fuck -the hell- up, you silly asshole, probably would be quite an
appropriate answer.
F.
G. Frege
(Stonelock) wrote:
>
> But i still do not agree that material implication
> does have semantic meaning.
>
Well, actually (and luckily) this does not depend on your agreement
concerning that question. :-)
It's like stating:
"But i still do not agree that the world is round."
(Stonelock II)
F.
G. Frege
wrote:
The following actually _is_ the "semantical meaning" of "->":
"A -> B" is false iff "A" is true and "B" is false;
it is true otherwise.
F.
Try to figure out yourself what this means in the case "q -> q".
G. Frege
(Stonelock) wrote:
>
> The problem is that propositional implication allows for any 2
> non-linked propositions to be evaluated on the same level while in
> fact, i think the truth value of the second argument is dependant on
> that of the first.
>
???
So you suggest that
(1 = 1) -> (1 = 1)
can't hold, since "the truth value of the second argument is dependant
on that of the first".
*sigh*
F.
tc...@lsa.umich.edu
>So would I. If you can prove it, I'll turn into a stuffed monkey right
>away. Of course that won't happen so....:).
So if "q -> q" makes sense, then "q -> q" makes sense. It's of course
ridiculous that something would cause itself, just as it would be ridiculous
for a proof to cause someone to turn into a stuffed monkey, but it's not
going to happen so....:)
Naturally, if you're going to be stubborn and refuse to admit you're wrong,
then you're going to be stubborn and refuse to admit you're wrong....:)
George Dance
In propositional implication, the truth value of the second assertion
(the consequent) is dependent on that of the first (the antecedent).
(P->Q) says that if P is true, then Q is true.
Similarly, (Q->Q) says that if Q is true, then Q is true. Is *that*
assertion true or not? If Q is true, is Q true or not?
Dennis E. Hamilton
P Q P->Q
F F T
F T T
T F F
T T T
This bothers some people and they use later versions of logic where implication is not this "material implication".
It seems to me that most of the problem is assuming that the truth of P->Q somehow imply the truth of Q by itself. It does not.
The classical P -> Q is equivalent to ~P or Q, so Q -> Q is equivalent to ~Q or Q and that is tautologously true in classical symbolic logic.
What is one to make of this notion of P -> Q? Well, it has to do with the principle of modus ponens as I was fortunate enough to have someone demonstrate to me.
Modus Ponens is a (classical) inference principle that takes the following form:
1. If it is established that P is true.
2. If it is established that P -> Q is true (this says very little about the particular values that P and Q have at the moment, though it does rule out one possibility)
3. Then one may infer that Q must be true.
Notice that Q -> Q is perfectly appropriate in this case, even in the informal sense of implication. That is, if you have established that Q is true, you have established that Q -> Q is true and you have established therefore, that Q is true. No biggie. It is not telling you anything more than that. In particular, Q -> Q does not tell you that Q is or must be true. It just says that the truth of Q implies that Q is true! From this perspective, it would be surprising if it didn't.
Finally, in every row of the above table where P is true and P->Q is true, it is also the case that Q is true. So the inference rule of modus ponens works perfectly well with this definition of P->Q and the usual propositional-logic truth-table semantics. Just keep in mind that, for classical symbolic logic, the truth of an implication does not imply by itself anything about the truth (or not) of the two components other than P -> Q being true is not consistent with P being true and Q being false. And that's enough for everything to work.
-- orcmid
Dennis E. Hamilton
AIIM DMware Technical Coordinator
---------------------------------------------------------
dennis....@acm.org
http://www.dmware.org
ODMA Support: http://ODMA.info/
"George Dance" <georg...@hotmail.com> wrote in message news:6312c50b.02111...@posting.google.com...
> crypto_s...@hotmail.com (Stonelock) wrote in message news:<64f2b1e9.02111...@posting.google.com>...
> > tc...@lsa.umich.edu wrote in message news:<ar2ik2$i4f$2...@galois.mit.edu>...
> > > In article <64f2b1e9.02111...@posting.google.com>,
> > > Stonelock <crypto_s...@hotmail.com> wrote:
[ ... ]
George Dance
> On 16 Nov 2002 04:38:27 -0800, georg...@hotmail.com (George Dance)
> wrote:
> > 1. (bad smell)
> > 2. ((bad smell)->(bad food))
> > ------
> > 3. (bad food)
>
> There's an interesting "reverse" relation between "causation" and
> implication.
>
> If the bad food actually _causes_ the bad smell, denoted by:
>
> (bad smell) <<- (bad food)
>
> then the PC sentence/proposition
>
> (bad smell) -> (bad food)
>
> holds.
>
> In general:
>
> if A <<- B then A -> B. (*)
>
> Since B can not be the cause of A and (at the same "time") A be true,
> but B false.
That (and what you go on to say in the balance) is true of necessary
causation. OTOH, if we were discussing sufficient (but not necessary)
causation, the relation would be
if A ->> B then A -> B
but not the reverse. Eg, if I cleaned the fridge, that would be
sufficient to eliminate the smell but not necessary (my wife could
clean it instead). So it would not be valid to infer, from the fact
that the fridge was clean, that I had cleaned it.
Stonelock
I disagree perfectly with you. This is why I object! I do not think
implication captures the true notion of implication as presently used.
> > > I don't know of any systematic treatment that inteprets -> as
> > > causal implication, though I'm sure that someone must have
> > > come up with one. Actually that's probably an underestimate;
> > > likely there are many different competing systems with such
> > > a motivation over which their proponents have endless and
> > > incomprehensible arguments.
> >
> > I dont think any operator pretending implication should be anything
> > else than causal implication
>
> Then you misunderstand what implication means in PC. 'A implies B'
> means that if A is true, then B is also true;
No no no! Propositional implication is not so. This would be modus
ponens, not propositional implication. Propositional implication is a
binary operator, not a unary one. The implication doesn't care about
reference frames and as thus can allow for the evaluation of ANYTHING.
"Me disagreeing with you -> me eating dinner" Both can be true, and
the implication will yield true even though there is no link between
the first proposition and the second other than me.
so if you know A, you
> can also validly conclude B; and you know B, by the same evidence that
> you know A.
>
> To take the first example off the top of my head: I smell a bad smell
> in my refrigerator, so I conclude that some food has gone bad in it.
> I know the smell is there, so I know the bad food is there:
>
> 1. (bad smell)
> 2. ((bad smell)->(bad food))
> ------
> 3. (bad food)
This is modus ponens, not boolean implication as seen in First order
logic.
> That knowledge that there is bad food in my fridge gives me a reasoned
> justification for looking for it and throwing it out.
>
> Now you come along and tell me that that's all wrong; either I cannot
> make such an inference at all, or, if I can, I cannot do so logically:
> since (bad food) is the cause of (bad smell), and I am allowed to use
> logic to reason only from sufficient causes to effects, and not from
> effects to necessary causes.
>
> Just why should I accept that? I know that inferences like this are
> sound; why should I believe that they are not logical, just because
> you say they aren't?
>
> > ;) hehe.
>
> If I discovered a bad smell in your fridge, and told you there was
> some bad food in there; and your only response was "hehe"; why would
> that change my mind?
Hey, thats up to you, and you only :). If thats enough to change your
mind...
Stonelock
Stonelock
> In article <64f2b1e9.02111...@posting.google.com>,
> >So would I. If you can prove it, I'll turn into a stuffed monkey right
> >away. Of course that won't happen so....:).
>
> So if "q -> q" makes sense, then "q -> q" makes sense. It's of course
> ridiculous that something would cause itself, just as it would be ridiculous
> for a proof to cause someone to turn into a stuffed monkey, but it's not
> going to happen so....:)
MWhaha, so we agree then :).
> Naturally, if you're going to be stubborn and refuse to admit you're wrong,
> then you're going to be stubborn and refuse to admit you're wrong....:)
Exactly, it doesnt prove anything.
Stonelock
G. Frege
wrote:
>
> That (and what you go on to say in the balance) is true of necessary
> causation. OTOH, if we were discussing sufficient (but not necessary)
> causation, the relation would be
>
> if A ->> B then A -> B
>
Well...
I didn't want to make a differentiation between necessary and
sufficient _conditions_ here. (But rather talk about the notation of
_cause_.)
Imho, the important thing is, that in the case where B is the _cause_
of A, A -> B always holds. With other word's if B is not a necessary
condition of A it CAN'T be seen as _the_ cause of A (in the case A
actually is the case).
F.
If we have more than one (possible) cause(s), say B_1, B_2, B_3, for
A, then A -> (B_1 v B_2 v B_3) would hold.
tc...@lsa.umich.edu
>> In article <64f2b1e9.02111...@posting.google.com>,
>> >So would I. If you can prove it, I'll turn into a stuffed monkey right
>> >away. Of course that won't happen so....:).
>>
>> So if "q -> q" makes sense, then "q -> q" makes sense. It's of course
>> ridiculous that something would cause itself, just as it would be ridiculous
>> for a proof to cause someone to turn into a stuffed monkey, but it's not
>> going to happen so....:)
>
>MWhaha, so we agree then :).
Do we? When I said "So if `q -> q' makes sense, then `q -> q' makes sense,"
did that make sense to you? Sounds like it did. But that was an instance
of `p -> p' (where p = "`q -> q' makes sense").
>> Naturally, if you're going to be stubborn and refuse to admit you're wrong,
>> then you're going to be stubborn and refuse to admit you're wrong....:)
>
>Exactly, it doesnt prove anything.
Did what I say make sense to you? It was another instance of `p -> p'
(this time with p = "you're going to be stubborn and refuse to admit
you're wrong"). It doesn't *prove* anything, but nobody was claiming
that it *proved* anything, only that it makes sense.
How is that you twice managed to agree with something that makes no sense?
G. Frege
>
> How is that you twice managed to agree with something that makes no sense?
>
Well, it's the non-sense he agrees with. :-)
F.
R. Srinivasan
> mean anything semantically?
> A cause can never be its consequence!!!!!!
>
classsical logic, a (syntactic) proof of q->q is just q->q, period. So
there is no question of "how the hell can you prove... q->q..."; even
if it doesn't mean anything semantically as you allege, the syntactic
proof is just what I stated above. Your bland rejection of q->q
implies that you do not agree with classical logic. Do you accept the
law of the excluded middle, qv~q and its classical equivalence to
q->q?
In constructive logic, q->q is accepted. But the excluded middle qv~q
will be accepted if and only if you have a either proof of q or a
proof of ~q. So constructivists clearly do not consider q->q to be
equivalent to qv~q. The constructivist definitions are as follows:
To prove qv~q, you must either have a proof of q or a proof of ~q.
To prove p->q, you must have an algorithm that converts a proof of p
into a proof of q. From this, q->q follows.
In the logic NAFL that I have proposed (see my preprint
PITT-PHIL-SCI00000635), the equivalence of qv~q and q->q is accepted.
But neither of these is a legitimate proposition of a theory T when q
is T-undecidable, i.e., when you can't prove either q or ~q in T. In
NAFL, as in classical logic, q->q is just "If q (~q), then q (~q)".
But when q is undecidable in a consistent T, "If q (~q)..." is taken
as an *axiomatic declaration* of the truth of q (~q) in NAFL, and so
q->q is only a legitimate proposition of T+q or T+~q, but not T (in
fact q->q fails in T, since it is equivalent to qv~q, which NAFL
doesn't accept). Here I use T+q (T+~q) to denote any consistent
extension of T in which you can prove q (~q).
Sincerely,
R. Srinivasan srad...@in.ibm.com
G. Frege
wrote:
>
> You first have to specify what logic you are talking about. In
> classical logic, a (syntactic) proof of q->q is just q->q, period.
> So there is no question of "how the hell can you prove... q->q"
>
Well, in a usual system of natural deduction (for classical logic) one
probably would have the proof:
1 (1) q A
(2) q -> q 1 CP
Hence: q -> q is a theorem.
Not a big deal though.
> [and] even if it doesn't mean anything semantically as you allege,
> the syntactic proof is just what [is] stated above.
>
F.
George Dance
> In classical (!) symbolic logic, the truth of P -> Q is determined by
> the following truth table:
>
> P Q P->Q
> F F T
> F T T
> T F F
> T T T
>
> This bothers some people and they use later versions of logic where
> implication is not this "material implication".
>
> It seems to me that most of the problem is assuming that the truth of
> P->Q somehow imply the truth of Q by itself. It does not.
People do make this mistake; and this is compounded when they learn
that (P->Q) is true whenever P is false. For if the truth of (P->Q)
implies the truth of Q (which it doesn't), and the falsity of P
implies the truth of (P->Q) (which it does), then the falsity of P
would imply the truth of Q; IOW, Q would be true whenever P is false,
which is the opposite of what (P->Q) actually states.
> The classical P -> Q is equivalent to ~P or Q, so Q -> Q is equivalent
> to ~Q or Q and that is tautologously true in classical symbolic logic.
>
> What is one to make of this notion of P -> Q? Well, it has to do with
> the principle of modus ponens as I was fortunate enough to have someone
> demonstrate to me.
>
> Modus Ponens is a (classical) inference principle that takes the
> following form:
>
> 1. If it is established that P is true.
> 2. If it is established that P -> Q is true (this says very
> little about the particular values that P and Q have at the moment,
> though it does rule out one possibility)
> 3. Then one may infer that Q must be true.
>
> Notice that Q -> Q is perfectly appropriate in this case, even in the
> informal sense of implication. That is, if you have established that Q
> is true, you have established that Q -> Q is true and you have
> established therefore, that Q is true. No biggie. It is not telling
> you anything more than that. In particular, Q -> Q does not tell you
> that Q is or must be true. It just says that the truth of Q implies that
> Q is true! From this perspective, it would be surprising if it didn't.
Agreed. That was essentially my point below, as well.
> Finally, in every row of the above table where P is true and P->Q is
> true, it is also the case that Q is true. So the inference rule of
> modus ponens works perfectly well with this definition of P->Q and the
> usual propositional-logic truth-table semantics. Just keep in mind
> that, for classical symbolic logic, the truth of an implication does not
> imply by itself anything about the truth (or not) of the two components
> other than P -> Q being true is not consistent with P being true and Q
> being false. And that's enough for everything to work.
>
> -- orcmid
>
> Dennis E. Hamilton
> AIIM DMware Technical Coordinator
> ---------------------------------------------------------
> dennis....@acm.org
> http://www.dmware.org
> ODMA Support: http://ODMA.info/
>
>
> "George Dance" <georg...@hotmail.com> wrote in message
> news:6312c50b.02111...@posting.google.com...
> > crypto ston...@hotmail.com (Stonelock) wrote in message
> news:<64f2b1e9.02111...@posting.google.com>...
> > > tc...@lsa.umich.edu wrote in message
> news:<ar2ik2$i4f$2...@galois.mit.edu>...
> > > > In article <64f2b1e9.02111...@posting.google.com>,
George Dance
> georg...@hotmail.com (George Dance) wrote in message news:<6312c50b.02111...@posting.google.com>...
> > There's nothing wrong with your coming up with your own system of
> > logic dealing with physical causation. (You could start by checking
> > out one of the existing logics, such as dynamic logic, that does deal
> > with physical change.)
> >
> > If you wanted, you could even use all of the notation of PC in your
> > logic, as long as you defined your own language prcisely each time to
> > prevent confusion. If you did that, I doubt that anyone would be
> > objecting.
> >
> > What's causing objections, I think, is your insistence that everyone
> > else is using PC incorrectly, because you don't accept what they mean
> > by the terms they use; that PC should be used for what you want it to
> > be use, and should not be used for what everyone else who uses PC uses
> > it for.
snip
> > Do you understand what the notation means? My understanding is that
> > PC is the study of reasoning; the process by which people figure out,
> > given what little they know is true, what else is true. That notion
> > (which I hope is clear enough) is exactly the one captured by the PC
> > notion of implication.
>
> I disagree perfectly with you. This is why I object! I do not think
> implication captures the true notion of implication as presently used.
So what do you think the 'true notion of implication' actually is?
And more to the point, how would you express it in a truth table
(assuming that it is truth-functional)?
snip
> > Then you misunderstand what implication means in PC. 'A implies B'
> > means that if A is true, then B is also true;
>
> No no no! Propositional implication is not so. This would be modus
> ponens, not propositional implication.
No; modus ponens is the inference that "B is true" from the premises
"A is true" *and* "if A is true, then B is also true." MP is one
possible inference using that second statement . Another is modus
tollens: inferring "A is not true" from "B is not true" and "if A is
true, then B is also true." Different inferences, using the same
implication statement.
> Propositional implication is a
> binary operator, not a unary one.
Yes.
> The implication doesn't care about
> reference frames and as thus can allow for the evaluation of ANYTHING.
> "Me disagreeing with you -> me eating dinner" Both can be true, and
> the implication will yield true even though there is no link between
> the first proposition and the second other than me.
That's only true until the first time you disagree with me at a time
that you're not eating dinner. Implication statements can be easily
disproved.
If you want to have an even firmer link between propositions, you can
add a relatedness operator R(A,B) into your system, such that A->B is
true or false as in the classical tt when R(A,B) is true, and false
otherwise.
Of you can make up your own causal operator, as has been suggested,
and axiomatize whatever you want regarding implication from it.
The point is, there are lots of ways to capture the notion of
causality, in a system of logic, that do not require redefining
implication.
OTOH, if you really feel you have to redefine the material
conditional: what would you redefine it as? What values of q are you
proposing such that q->q is not a valid wff?
> > so if you know A, you
> > can also validly conclude B; and you know B, by the same evidence that
> > you know A.
> >
> > To take the first example off the top of my head: I smell a bad smell
> > in my refrigerator, so I conclude that some food has gone bad in it.
> > I know the smell is there, so I know the bad food is there:
> >
> > 1. (bad smell)
> > 2. ((bad smell)->(bad food))
> > ------
> > 3. (bad food)
>
> This is modus ponens, not boolean implication as seen in First order
> logic.
2 is an implication statement. 1-2-3 is a MP inference, that uses an
implication statement. There's a difference.
It sounds as though you believe that MP is a valid inference, which
means that you should consider its corresponding conditional,
(P&(P->Q))->Q, to be a valid wff.
Now, that is puzzling. What truth table values can you assign to
(P->Q), for any P and Q, such that (P&(P->Q))->Q is a valid wff, and
Q->Q is not?
I am really curious as to exactly what you are proposing.
George Dance
snip
> So his "question" actually is:
>
> "How the hell can ~q v q?????"
>
> Fuck -the hell- up, you silly asshole, probably would be quite an
> appropriate answer.
> F.
Appropriate for sci.math, maybe; but sci.logic should not be a flame
list, should it? 8)
Paul Holbach
> How the hell can you prove syntaxically that q -> q when it doesnt
> mean anything semantically?
> A cause can never be its consequence!!!!!!
Logical implication has nothing to do with causation. The abstract
principle of self-implication as expressed by p -> p is, first of all,
a consequence of the law of excluded middle, stating that it is
impossible that both p and not p.
It would really be absurd if one could conclude from the fact that it
is raining that it is not raining. The principle of self-implication
doesn´t convey any new information, but it makes sure that the world
of logic is intact; and that´s why it´s extremely valuable and far
from being "meaningless"...!
regards
PH
Thad Coons
news:881c8779.02111...@posting.google.com...
> crypto_s...@hotmail.com (Stonelock) wrote in message
news:<64f2b1e9.02111...@posting.google.com>...
>
> > How the hell can you prove syntaxically that q -> q when it doesnt
> > mean anything semantically?
> > A cause can never be its consequence!!!!!!
>
> Logical implication has nothing to do with causation. The abstract
> principle of self-implication as expressed by p -> p is, first of all,
> a consequence of the law of excluded middle, stating that it is
> impossible that both p and not p.
I would agree with the first sentence here, but disagree with the second
except in the special case of classical two-valued logic. The principle of
self-implication is more general and may continue to hold even when the
excluded middle fails.
Thad Coons
G. Frege
wrote:
> >
> > Fuck -the hell- up, you silly asshole, probably would be quite an
> > appropriate answer.
> >
>
> Appropriate for sci.math, maybe; but sci.logic should not be a flame
> list, should it? 8)
>
That's certainly a reasonable position, I agree. :-)
F.
R. Srinivasan
> On 17 Nov 2002 21:53:48 -0800, srad...@in.ibm.com (R. Srinivasan)
> wrote:
>
> >
> > You first have to specify what logic you are talking about. In
> > classical logic, a (syntactic) proof of q->q is just q->q, period.
> > So there is no question of "how the hell can you prove... q->q"
> >
> Well, in a usual system of natural deduction (for classical logic) one
> probably would have the proof:
>
> 1 (1) q A
> (2) q -> q 1 CP
>
> Hence: q -> q is a theorem.
>
> Not a big deal though.
>
priniple of classical logic that it must necessarily be stated as an
axiom in some form or the other. Of course, the actual proof will
depend on what the axioms are. If my understanding of your proof is
correct, you have stated that one could conclude (by CP) q from the
assumption q; hence q->q. I could turn this around and say that (in an
alternative system): From the assumption q, one could conclude q if
and only if you presume the validity of q->q (which must be stated as
an axiom in this system).
Sincerely,
R. Srinivasan srad...@in.ibm.com
G. Frege
wrote:
> >
> > In a usual system of natural deduction (for classical logic) one
> > probably would have the proof:
> >
> > 1 (1) q A
> > (2) q -> q 1 CP
> >
> > Hence: q -> q is a theorem.
> >
> > Not a big deal though.
> >
> You are probably right. I had assumed that q->q is such a fundamental
> principle of classical logic that it must necessarily be stated as an
> axiom in some form or the other.
>
Well, one could think so; actually this is not the case for _most_
("famous") axiom systems for propositional logic (in Hilbert-style).
>
> Of course, the actual proof will depend on what the axioms are.
>
Sure. For example in the Begriffsschrift we have the proofs:
Formula 26:
a -> b -> a from A1 (formula 1)
(a -> b -> a) -> b -> a -> a from A3 (formula 8)
---------------------------- Conclusion
b -> a -> a
Formula 27:
(a -> b -> a) -> a -> a from formula 26
a -> b -> a from A1 (formula 1)
---------------------------- Conclusion
a -> a
>
> If my understanding of your proof is correct, you have stated that
> one could conclude (by CP) q from the assumption q; hence q->q.
>
Yes. ( Well, actually: hence |- q->q. )
F.
G. Frege
wrote:
>
> For example in the Begriffsschrift we have the proofs:
>
> Formula 26:
>
> a -> b -> a from A1 (formula 1)
> (a -> b -> a) -> b -> a -> a from A3 (formula 8)
> ---------------------------- Conclusion
> b -> a -> a
>
> Formula 27:
>
> (a -> b -> a) -> a -> a from formula 26
> a -> b -> a from A1 (formula 1)
> ---------------------------- Conclusion
> a -> a
>
We could transform that into a single proof:
a -> (a -> b -> a) -> a from A1
(a -> (a -> b -> a) -> a) -> (a -> b -> a) -> a -> a from A3
----------------------------------------------------
(a -> b -> a) -> a -> a Concl.
a -> b -> a from A1
----------------------------------------------------
a -> a Concl.
Or in a familiar notation:
(1) q->((q->(p->q))->q) A1
(2) (q->((q->(p->q))->q))->((q->(p->q))->(q->q)) A3
(3) (q->(p->q))->(q->q) 2,1 MP
(4) q->(p->q) A1
(5) q->q 3,4 MP
F.
R. Srinivasan
> > You are probably right. I had assumed that q->q is such a fundamental
> > principle of classical logic that it must necessarily be stated as an
> > axiom in some form or the other.
> >
> Well, one could think so; actually this is not the case for _most_
> ("famous") axiom systems for propositional logic (in Hilbert-style).
>
Point taken. But consider the axiom system given in
<http://plato.stanford.edu/entries/logic-intuitionistic/#IQC>
which the author calls a "Hilbert-style formalism, from Kleene [1952],
for intuitionistic first-order predicate logic." The author notes that
upon replacing the axiom ~A->(A->B) by ~~A->A (or equivalently, by
Av~A), "a formal system for classical first-order predicate logic
results". Now this system includes the following axioms (among
others):
(1) A&B -> A
(2) A&B -> B
(3) A -> AvB
(4) B -> AvB
where A and B are any wff's.
It seems to me that A->A is trivially contained in any of (1)--(4). By
the way, I have seen these axioms attributed to Hilbert elsewhere on
the web.
Sincerely,
R. Srinivasan srad...@in.ibm.com
G. Frege
wrote:
> ...consider [...]
>
> (1) A&B -> A
> (2) A&B -> B
> (3) A -> AvB
> (4) B -> AvB
>
> where A and B are any wff's.
>
> It seems to me that A->A is trivially contained in any of (1)--(4).
>
No, not really, actually this axioms "encode" the usual "meaning" of &
and v.
If A&B is true we certainly would agree that A alone is true too, and
that B alone is true too.
On the other hand, if A is true and/or B is true, we certainly will
agree that AvB is true too.
Actually the axiom system you mentioned has the following first two
axioms (beside others):
A1 A -> (B -> A).
A2 (A -> B) -> ((A -> (B -> C)) -> (A -> C)).
And hence we can (and must) prove: A->A.
(1) A -> (A -> A) A1
(2) (A -> (A -> A)) -> ((A -> ((A -> A) -> A)) -> (A -> A)) A2
(3) (A -> ((A -> A) -> A)) -> (A -> A) 2,1 MP
(4) A -> ((A -> A) -> A) A1
(5) A -> A 3,4 MP
> By the way, I have seen these axioms attributed to Hilbert elsewhere on
> the web.
>
Well, they are not exactly the same, but it's a similar system. The
idea is that for any connective you have axioms that "encode" the
relevant properties.
Axiom System for propositional logic due to Hilbert-Bernays [1934]:
I. Implication:
p -> (q -> p)
(p -> (p -> q)) -> (p -> q)
(p -> q) -> ((q -> r) -> (p -> r))
II. Conjunction
p & q -> p
p & q -> q
(p -> q) -> ((p -> r) -> (p -> q & r))
III. Disjunction
p -> p v q
q -> p v q
(p -> r) -> ((p -> r) -> (p v q -> r))
IV. Equivalence
(p <-> q) -> (p -> q)
(p <-> q) -> (q -> p)
(p -> q) -> ((q -> p) -> (p <-> q))
V. Negation
(p -> q) -> (~q -> ~p)
p -> ~~p
~~p -> p
---------------------------------
As you can imagine, this is a quite convenient system. But still q->q
have to be proven:
(1) q -> (q -> p) I.1
(2) (q -> (q -> q)) -> (q -> q) I.2
(3) q -> q 2,1 MP
F.
G. Frege
wrote:
2 typos:
>
> (p -> r) -> ((q -> r) -> (p v q -> r))
>
...
>
> (1) q -> (q -> q) I.1
>
...
R. Srinivasan
Thank you for the clarifications. There is no doubt that the axiom
scheme A&B->A, for example, includes within it A&A->A. But to go from
A&A->A to A->A would still require proof, I suppose. Can we get around
this by imposing a metalogical substitution rule that all instances of
A&A (and AvA) in any wff can be replaced by A? A->A would then follow
trivially from A&A->A. One might argue that such a substitution rule
would presuppose A->A, which is what we are trying to prove, but it
actually doesn't. In my logic NAFL, A->A does not hold in general
(when A is T-undecidable in a consistent theory T), but such a
substitution rule would still be valid.
Sincerely,
R. Srinivasan srad...@in.ibm.com
G. Frege
wrote:
> Can we get around this by imposing a metalogical substitution rule
> that all instances of A&A (and AvA) in any wff can be replaced by A?
>
Yes of course.
Actually in Kleene's axiom system we have the additional axioms:
A & B -> A
A & B -> B
A -> (B -> A & B)
Hence we have the theorem:
A & A <-> A ,
[ if we define X <-> Y to mean (X -> Y) & (Y -> X) ].
And it's quite easy (=it can be done) to justify a meta-rule that
allows the replacement of A in a formula C by B, if A <-> B holds.
(Principle of replacement).
>
> A->A would then follow trivially from A&A->A.
>
Yes.
I have to admit (concerning Kleene's system): the more I look at it,
the more I like it! :-)
For the propositional calculus part, we have:
A -> (B -> A)
(A -> B) -> ((A -> (B -> C)) -> (A -> C))
A & B -> A, A & B -> B
A -> (B -> A & B)
A -> A v B, B -> A v B
(A -> C) -> ((B -> C) -> (A v B -> C))
~A -> (A -> B)
(A -> B) -> ((A -> ~B) -> ~A)
Kleene [1952].
In this system the _introduction_ and _elimination_ of connectives is
even "cleaner" (sic!) realized then in Hilbert-Bernays' system;
actually I don't think it can be done (much) "better".
The _effect_ of this axioms can be made quite obvious by using an
alternative notation:
->-introduction:
A -> (B -> A)
A
-------
B -> A
Given A we may derive B -> A.
->-elimination:
(A -> B) -> ((A -> (B -> C)) -> (A -> C))
A -> (B -> C); A -> B
-----------------------
A -> C
Under the condition A -> B we can derive A -> C from A -> (B -> C).
&-elimination:
A & B -> A, A & B -> B
A & B A & B
------- --------
A B
&-introduction:
A -> (B -> A & B)
A , B
-------
A & B
Given A and B we may derive A & B.
v-introduction:
A -> A v B, B -> A v B
A B
------- --------
A v B A v B
Given A and/or given B we may derive A & B.
v-elimination:
(A -> C) -> ((B -> C) -> (A v B -> C))
A v B -> C; A -> C, B -> C
-------
C
Under the conditions A -> C and B -> C we can derive C from
A v B -> C.
~-elimination:
~A -> (A -> B)
A, ~A
-------
B
Given A and ~A we can derive B.
From the axioms A & B -> A, A & B -> B
we immediately have, that from A & ~A we
can derive A and ~A. Hence we have the
well known principle here, that from a
contradiction anything can be derived.
~-introduction:
(A -> B) -> ((A -> ~B) -> ~A)
A -> B, A -> ~B
-------
~A
Given A -> B and A -> ~B we can derive ~A.
If a condition A leads to B and also leads to ~B (and hence to a
contradiction) we can conclude ~A.
Great system!
F.
G. Frege
made a typo:
Should read:
>
> v-introduction:
> ...
> Given A and/or given B we may derive A v B. <----
>
G. Frege
>
> Actually in Kleene's axiom system we have the additional axioms:
>
> A & B -> A
> A & B -> B
> A -> (B -> A & B) (&I)
>
> Hence we have the theorem:
>
> A & A <-> A ,
>
> [ if we define X <-> Y to mean (X -> Y) & (Y -> X) ].
Hmmm... probably I was a little bit to fast... here.
Clearly from the axioms above we have:
A & A -> A.
But how to derive the other way round
A -> A & A ?
Well, together with axiom 1 and axiom 2
A1 A -> (B -> A).
A2 (A -> B) -> ((A -> (B -> C)) -> (A -> C)).
this can be proved:
(1) A -> (A -> A) A1
(2) (A -> (A -> A)) -> ((A -> ((A -> A) -> A)) -> (A -> A)) A2
(3) (A -> ((A -> A) -> A)) -> (A -> A) 2,1 MP
(4) A -> ((A -> A) -> A) A1
(5) A -> A 3,4 MP
(6) (A -> A) -> ((A -> (A -> A & A)) -> (A -> A & A)) A2
(7) (A -> (A -> A & A)) -> (A -> A & A) 6,5 MP
(8) A -> (A -> A & A) &I
(9) A -> A & A 7,8 MP
Note that as an intermediate result in line 5 we have again A -> A.
Actually we conclude in the following way from Axiom 2:
If we have A -> A' and also have A -> (A' -> A & A), then we
can conclude A -> A & A. In our case we just have A' = A.
--------------------------------------
Hence _finally_ we have shown:
A & A <-> A
[ with X <-> Y to mean (X -> Y) & (Y -> X) ].
Now using the proposed meta-rule (of replacement) we can use this
result to get from the axiom
A & A -> A
the formula
A -> A
( by replacing "A & A" with "A").
Well, but as you see, we could have this "cheaper". ;-)
F.
Stonelock
> In article <64f2b1e9.02111...@posting.google.com>,
> Stonelock <crypto_s...@hotmail.com> wrote:
> >tc...@lsa.umich.edu wrote in message news:<ar5qa3$hu0$1...@galois.mit.edu>...
> >> In article <64f2b1e9.02111...@posting.google.com>,
> >> >So would I. If you can prove it, I'll turn into a stuffed monkey right
> >> >away. Of course that won't happen so....:).
> >>
> >> So if "q -> q" makes sense, then "q -> q" makes sense. It's of course
> >> ridiculous that something would cause itself, just as it would be ridiculous
> >> for a proof to cause someone to turn into a stuffed monkey, but it's not
> >> going to happen so....:)
> >
> >MWhaha, so we agree then :).
>
> Do we? When I said "So if `q -> q' makes sense, then `q -> q' makes sense,"
> did that make sense to you? Sounds like it did. But that was an instance
> of `p -> p' (where p = "`q -> q' makes sense").
Jesus, give me a lil credit here. In all good spirit i thought you
were having a good laugh seeing what i had brought up.
> >> Naturally, if you're going to be stubborn and refuse to admit you're wrong,
> >> then you're going to be stubborn and refuse to admit you're wrong....:)
> >
> >Exactly, it doesnt prove anything.
>
> Did what I say make sense to you?
YEAHS IT SHOWED THAT IT DIDNT MAKE SENSE, IT DIDNT PROVE ANYTHING.
It was another instance of `p -> p'
> (this time with p = "you're going to be stubborn and refuse to admit
> you're wrong"). It doesn't *prove* anything, but nobody was claiming
> that it *proved* anything, only that it makes sense.
>
> How is that you twice managed to agree with something that makes no sense?
bah
Stonelock
G. Frege
wrote:
>
> Axiom System for propositional logic due to Hilbert-Bernays [1934]:
>
Note that for each connective we have axioms for "introduction" and
"elimination".
> I. Implication:
>
> p -> (q -> p)
> (p -> (p -> q)) -> (p -> q)
> (p -> q) -> ((q -> r) -> (p -> r))
>
Introduction: Given p, with p -> (q -> p) we can derive q -> p.
Elimination: Given p -> (p -> q), with (p -> (p -> q)) -> (p -> q) we
can derive p -> q.
Chain rule: Given p -> q and q -> r with (p -> q) -> ((q -> r) -> (p
-> r)) we can derive p -> r.
> II. Conjunction
>
> p & q -> p
> p & q -> q
> (p -> q) -> ((p -> r) -> (p -> q & r))
>
Elimination: Given p & q, with p & q -> p we can derive p. Given p &
q, with p & q -> q we can derive q.
Introduction: Given p -> q and p -> r, with (p -> q) -> ((p -> r) ->
(p -> q & r)) we can derive p -> q & r.
> III. Disjunction
>
> p -> p v q
> q -> p v q
> (p -> r) -> ((q -> r) -> (p v q -> r))
>
Introduction: Given p, with p -> p v q we can derive p v q. Given q,
with q -> p v q we can derive p v q.
Elimination: Given p v q, p -> r and q -> r, with (p -> r) -> ((q ->
r) -> (p v q -> r)) we can derive r.
> IV. Equivalence
>
> (p <-> q) -> (p -> q)
> (p <-> q) -> (q -> p)
> (p -> q) -> ((q -> p) -> (p <-> q))
>
Elimination: Given p <-> q, with (p <-> q) -> (p -> q) we can derive
p -> q. Given p <-> q, with (p <-> q) -> (q -> p) we can derive q ->
p.
Introduction: Given p -> q and q -> p, with (p -> q) -> ((q -> p) ->
(p <-> q)) we can derive p <-> q.
> V. Negation
>
> (p -> q) -> (~q -> ~p)
> p -> ~~p
> ~~p -> p
>
Introduction: Given p -> q, with (p -> q) -> (~q -> ~p) we can derive
~q -> ~p. Given p, with p -> ~~p we can derive ~~p.
Elimination: Given ~~p, with ~~p -> p we can derive p.
--------------------------------------
Imho, the resulting axiom system is not without a certain appeal.
F.
G. Frege
wrote:
>
> Axiom System for propositional logic due to Hilbert-Bernays [1934]:
>
Note that their system is partly identical with Frege's system, to be
found in Begriffsschrift:
F1 a -> (b -> a)
F2 (a -> (b -> c)) -> ((a -> b) -> (a -> c))
F3 a -> (b -> c) -> (b -> (a -> c))
F4 (a -> b) -> (~b -> ~a)
F5 ~~a -> a
F6 a -> ~~a
Since F3 can be proven from F1, F2 alone, the following axiom system
would suffice for PC:
F1 a -> (b -> a)
F2 (a -> (b -> c)) -> ((a -> b) -> (a -> c))
F4 (a -> b) -> (~b -> ~a)
F5 ~~a -> a
F6 a -> ~~a
Hilbert-Bernays now obviously (a) just replaced F2 by:
I.2 (p -> (p -> q)) -> (p -> q)
I.3 (p -> q) -> ((q -> r) -> (p -> r))
The first of which seems to be a "novelty" that can be attributed to
Hilbert-Bernays. The second is the well known "rule" of "backward
propagation" [ see Lukasiewicz, 1929 ]
And (b) added some axioms governing the connectives &, v, <->. (Note
that Frege only had "->" and "~" as primitive "connectives" in his
system.)
>
> I. Implication:
>
> p -> (q -> p) [ F1 ]
> (p -> (p -> q)) -> (p -> q) |- replacing [ F2 ]
> (p -> q) -> ((q -> r) -> (p -> r)) |
>
> II. Conjunction
>
> p & q -> p
> p & q -> q
> (p -> q) -> ((p -> r) -> (p -> q & r))
>
> III. Disjunction
>
> p -> p v q
> q -> p v q
> (p -> r) -> ((p -> r) -> (p v q -> r))
>
> IV. Equivalence
>
> (p <-> q) -> (p -> q)
> (p <-> q) -> (q -> p)
> (p -> q) -> ((q -> p) -> (p <-> q))
>
> V. Negation
>
> (p -> q) -> (~q -> ~p) [ F4 ]
> p -> ~~p [ F6 ]
> ~~p -> p [ F5 ]
>
Now from Frege's system we had:
F1 a -> (b -> a)
F2 (a -> (b -> c)) -> ((a -> b) -> (a -> c))
F4 (a -> b) -> (~b -> ~a)
F5 ~~a -> a
F6 a -> ~~a
If we replace F4, F5, F6 just by
A3 (~a -> ~b) -> (b -> a) ,
we get the famous system [ Lukasiewicz & Tarski, 1930 ]:
A1 a -> (b -> a)
A2 (a -> (b -> c)) -> ((a -> b) -> (a -> c))
A3 (~a -> ~b) -> (b -> a)
F.
R. Srinivasan
as Kleene's axioms. I plan to study these systems more closely for the
purposes of my own research.
Sincerely,
R. Srinivasan srad...@in.ibm.com
Stonelock
Objectively, there is only one logic that stands; frames of reference
logic. The rest are poor derivates and should not be called logic.
Stonelock