General Question.

[Bryan]

is {V : A & B} / {V <-- Q} the same as just A & B?

[Behnam Robatmili]

No, I don't think it is the case! First, I have no idea what {V : A &
B} means. By definition, A inter B is equal to {v: (v in A) & (v in
B)}. Second, {V : A & B} / {V <-- Q} is equal to {Q : A & B}. Right?
It also doesn't make sense.

Behnam

[dsny...@gmail.com]

I'm not sure if this is what he was getting at, but on the problems on
the HW, I don't know what to do with the set notation when there isn't
a value to call member on and remove it. If you have

{v: (v in A) | (v in B)}

do you just treat it like a quantifier and drop the brackets to let v
be the variable, since you don't have a substitution?

[Behnam Robatmili]

No, you can't do that because none of the set operations allows you to
introduce new variable. The only rule that lets you somehow drop the
brackets is the member rule.
Here is its simple form:
(e in {v: (P v)}) <-> (P e)

Behnam

[Bryan]

What I was more or less getting it, was that if you have {v : (v in A) &
(v in B)}) That you could just reduce it to A & B by letting A be psia
& B be psib, doing a replace on both psia and psib of {v <-- Q}, which
then reduces to and then doing the substitution {v <-- Q}, and since v
does not exist in psi, then it's just psi, which is just A & B.

[Bryan]

Or, what about {Q : A & ({Q : B & C}) == {Q : A & B & C}.

[Behnam Robatmili]

Bryan,

By A & B, do you mean A inter B or do you really mean anding A and B?
Because if you mean A inter B, then according to the intersection
rule {v : (v in A) & (v in B)} is equal to A inter B.

Behnam

[Behnam Robatmili]

I'm slightly confused by your notation. I think by {Q : A & ({Q : B &
C})}, you mean sth like {v: v in (A inter {v: v in (B inter C)})}. Is
that right?

Behnam

[Ian W.]

{v : v in A & v in B} is a set, which represents an object; a thing. A
& B is a formula, which represents a proposition; a true-or-false
statement. So these are certainly not the same. Nor can you use any of
the inference rules to prove that they are the same.

Ian

On Apr 26, 7:20 pm, Behnam Robatmili <be...@cs.utexas.edu> wrote:
> I'm slightly confused by your notation. I think by {Q : A & ({Q : B &  
> C})}, you mean sth like {v: v in (A inter {v: v in (B inter C)})}. Is  
> that right?
>
> Behnam
>
> On Apr 26, 2009, at 5:16 PM, Bryan wrote:
>
>
>
> > Behnam Robatmili wrote:
> >> No, you can't do that because none of the set operations allows you  
> >> to
> >> introduce new variable. The only rule that lets you somehow drop the
> >> brackets is the member rule.
> >> Here is its simple form:
> >> (e in {v: (P v)}) <-> (P e)
>
> >> Behnam
>

[Bryan]

What I mean is that, by the definition of Intersection, A intersection
(B intersection C) becomes {V : (V in A) and (V in ({V : (V in B) and
(V in C)})}
Which, if you use the propositional equivalence of {v : psi} / {v <--
e}, what I am asking is: Do you get {V A and ({ V : B and C})} And if
so, can you combine it to {V : A and B and C}. If not, then what does
it reduce to?

On Apr 26, 7:20 pm, Behnam Robatmili <be...@cs.utexas.edu> wrote:
> I'm slightly confused by your notation. I think by {Q : A & ({Q : B &  
> C})}, you mean sth like {v: v in (A inter {v: v in (B inter C)})}. Is  
> that right?
>
> Behnam
>
> On Apr 26, 2009, at 5:16 PM, Bryan wrote:
>
>
>
> > Behnam Robatmili wrote:
> >> No, you can't do that because none of the set operations allows you  
> >> to
> >> introduce new variable. The only rule that lets you somehow drop the
> >> brackets is the member rule.
> >> Here is its simple form:
> >> (e in {v: (P v)}) <-> (P e)
>
> >> Behnam
>

[Behnam Robatmili]

Ok. I see your point.

{v : (v in A) & (v in ({v : (v in B) & (v in C)})} is a set and you
cannot change it to {v: A and {v: in B and C}}. You only use member
rule when you have a membership. So if you had e in {v : (v in A) & (v
in ({v : (v in B) & (v in C)})} then it would be propositionally
equivalent to e (e in A) & (e in ({v : (v in B) & (v in C)}). Now you
could apply that rule again and get: ...

Does that makes sense?

Behnam