Is this quantifier statement $(\forall x)(\phi (x) \implies \psi(x)) \implies ((\forall x)\phi(x)\implies (\forall x)\psi (x)) $ always true?

Question

I have question whether I am solving this problem correctly I am checking if this statement is always true $(\forall x)(\phi (x) \implies \psi(x)) \implies ((\forall x)\phi(x)\implies (\forall x)\psi (x)) $

I tried assuming that it can be false , so $((\forall x)\phi(x)\implies (\forall x)\psi (x))$ =0 and then $(\forall x)(\phi (x) \implies \psi(x)) = 1 $. Opening up the second one i get that it can be = zero only when $(\forall x)\phi(x)=1$ and $(\forall x)\psi (x)) = 0 $. Hence the whole statement can`t be false. Am i right?

Answer 1

You are almost there, and your intuition is correct. You still need to show why the condition which makes the statement false, never happens.

Assume for the purpose of contradition:

$(\forall x)\psi (x) = 0 $

Then, there must be atleast one witness for which $\psi(x)$ does not hold. ( in essence, there must be a counterexample)

$(\exists x)(\neg(\psi (x))) = 1 $

Pick ξ so that $\neg(\psi (ξ)) = 1 $

But, we are considering the case which makes

note: I am assuming the OP has a typo in the question, I filled in brackets where I felt appropriate.

$[(\forall x)((\phi (x) \implies \psi(x))] \implies [(\forall x)\phi(x)\implies (\forall x)\psi (x)] $

false

Which, requires ( as you rightly noticed)

$(\forall x)((\phi (x) \implies \psi(x))$

$(\forall x)(\phi (x)$

to both be true.

Since both are Universally Quantified.

Subsitute our counterxample ξ

$\phi (ξ) \implies \psi(ξ) $

$\phi (ξ)$

by Modus Ponens

$\psi(ξ) $ = 1

Which is our desired contradiction.