# How to prove equivalence when introducing quantifiers?

Given that $P(x) \Leftrightarrow Q(x)$, I want to check if the followings are correct?

• $\forall x. P(x) ~ \Leftrightarrow ~ \forall x. Q(x)$
• $\exists x. P(x) ~ \Leftrightarrow ~ \exists x. Q(x)$

I think that they are correct, but I do not know how to formally prove it.

Can anyone help me to prove or disprove it?

Thank you very much!

• What does $P(x)\Leftrightarrow Q(x)$ mean, when you haven't specified $x$? Jul 19, 2015 at 8:24
• Hi, here $x$ is a variable and $P(x)$ and $Q(x)$ are two predicates Jul 19, 2015 at 8:57

You may prove these either by using truth trees, or by contradiction. I will leave out a few steps

I)$\forall x P(x)$ And this implies

II)$\forall x \not Q(x)$

Use any $a$ to instantiate ,

III)$\not Q(a)$ Now use I) , to conclude.....for this $a$:

IV) $P(a)$ Now use:

V) $P(x) \Leftrightarrow Q(x)$ (specifically, use that $P(x) \Leftrightarrow Q(x)$ implies $P \Rightarrow Q$ and $Q \Rightarrow P$) , to derive a contradiction.

Let me know if you want to use truth trees.

• Hi, thanks for your efforts! It seems that your proof shows that $\forall x. P(x) \not\Rightarrow \neg (\forall x. Q(x))$. Is it equivalent with $\forall x. P(x) \wedge \forall x. Q(x)$? Can you show how to use the truth tree? Thank you very much! Jul 19, 2015 at 9:01
• @TrungTa: It is a proof by contradiction: I assume $A$, separately assume not $B$, arrive at a contradiction, so I conclude $B$, and then I conclude $A$ implies $B$. Jul 19, 2015 at 14:57
• Thanks for your clarification. Yes, you are right! In addition, I think that the words "∀xP(x) and this implies ∃xQ̸(x)" in your proof confuses me. May be your assumption is just "∀xP(x) and ∃xQ̸(x)" are Valid. (as mentioned in your comment). Thank you very much, again! Jul 19, 2015 at 15:11
• @TrungTa: My apologies, that was a typo. Please see the edits and let me know. Jul 19, 2015 at 15:27