Question 270

[Liz]

Gamma: (all x ((p x) --> (q x y)))
Gamma': (all x ((p 3) --> (q 3 y)))

Define p and q so that Gamma is always true but Gamma' is not always
true. Assume the Domain consists of natural numbers.

I am really stuck on this once because Gamma basically translates to :
For all natural numbers, p(x) implies q(x,y). In order for that to be
true:
a) p(x) is always false and q(x,y) can be anything OR
b) p(x) is always true and q(x,y) is always true.

If that is so, then isn't it impossible for there to be a function p
and q which will make Gamma' false?
If I take route a, then Gamma' will always be true since Nil--->
anything <--> True.
If I take route b, then Anything ---> T <--> True.

For Gamma' to be Invalid, it seems that p must always return True, but
q can return either nil or true. If that is so, it would not be True
for all x (Gamma) since if I chose x = 3 then it should have to return
a false statement.

Does anyone understand what I am talking about?

[A.J. Gardner]

You have written gamma and gamma' incorrectly: read the parentheses carefully.

((all x (p x)) --> blah x blah blah)

is distinct from

(all x ((p x) --> blah x blah blah)).

Cheers,

[Liz]

Oh wow that makes a lot more sense now thank you =]

[Hyon Choe]

me and my friend are still very confused on this one...
can anyone explain further?..

[Leo Schnee]

Keep in mind that nil --> anything true.
That may help you a bit.
If you still need help:
I defined P to only be true when X is 3.
Making gamma be nil --> anything,
and gamma' t --> (q 3 y).

I believe that this is how you do it. Correct me if I am wrong.

-Leo