Q 350

[Anthony McKnight]

Wow... any way I go down this one, it looks buggered. Any suggestions for how to attack this one?

[Ian W.]

I suggest starting with the formal definitions of the power set
operator and the subset relation. You might also compute some small
examples (e.g., A = {1, 2, 3} and B = {4, 5}) for some insight, and
also to convince yourself (or not) that the statement is actually
true.

On Apr 24, 7:04 pm, Anthony McKnight <anthonymckni...@gmail.com>
wrote:

[dsny...@gmail.com]

When you're going from this step to the step after powerSet has been
substituted, doesn't the ' v ' in {v: v sube A} need to be renamed,
since there is already a ' v ' variable floating around that is bound
by the quantifier?

(forall v: (v in powerSet(A)) --> (v in powerSet(A union B)))
= {powerset, twice}
(forall v: (v in {v: v sube A}) --> (v in {v: v sube (A union B)}))

[todd]

im having problems with this one too.

i get down to something like this:
(forall v: (v in (forall r: r in A))) --> (v in (s in ((p in A) || (p
in B))))

and i see that i have a universal hypothesis, but i dont know how
that's really going to help me.

> > > how to attack this one?- Hide quoted text -
>
> - Show quoted text -

[Aaron]

ok so I am stuck on this on e I have gotton to
(w subset A) -> ((w in A) || (w in B)) -------- which is now obviously
true but when I try and finish out the subset....
={def of subset}
(forall v: (v in w) -> (v in A)) -> ((w in A) || (w in B))
={for all hyp}
(v in w) -> (v in A) -> ((w in A) || (w in B)) --- now this looks
wrong. Just because v is in w and in A does not mean w is in A. and
that is what I need

What did I do wrong here?

[Nathan Wetzler]

You might need a theorem like 341 in your proof...