about counter-example for Quantifier based statement (logic)

Question

I have a logic based statement of this form:

$$ (\forall x\ ,p(x))\to (\exists y\ , q(x,y))) $$

I am trying to find out if this statement is TRUE or False. I have 2 methods of proof, and one leads to FALSE, the other leads to TRUE. I have a problem hence!

MY METHOD PART 1:

So for the first part of this logic equation I call it P and the second part I call it Q, so i get an overall implication equation:

$$ P \to Q $$

So I thought I could do a counter-example pick an x-value in order to set P to be TRUE and then pick a y-value to make Q-false. That way in an implication statement if P=TRUE and you have Q=FALSE, then overall the implication is FALSE which would disprove the overall statement.

I have seen on some other site that you can do this in order to disprove this.

BUT, I now have doubts.

MY METHOD PART 2:

Because in the first part it has the "For All" quantifier which in my notation is inside P. SO now i will take the "For ALL" into account. . I can show that the function inside P, p(x), I can find a counter-example x-value to show that the left side, capital P as FALSE.

NOW in an IMPLICATION if the first part P is FALSE, then based on truth-table for this, it does not matter if Q is TRUE or FALSE. Because as soon as P is False it makes the whole implication statement TRUE!!!

SO now I did 2 Methods, the first one shows that the overall statement is "FALSE" and I have a second Method that shows that the overall statement is "TRUE".

SO I am not sure which method is the problem.

Hope someone can clarify this.

Answer 1

The formula is neither true nor false in isolation.

It only gets a truth value when you select an interpretation to evaluate it, which fixes which universe the quantifiers range over and when the $p$ and $q$ predicates are true.

And even so, you also need to choose a value for the appearance of $x$ on the right-hand side of $\to$, which is not in scope of the $\forall x$.

Answer 2

I am trying to find out if this statement is TRUE or False. $$ (\forall x\ \,p(x))\to (\exists y\ \, q(x,y)) $$

The given statement is false under this interpretation (your Method 2): \begin{align}\text{universe of discourse}&:=\mathbb R\\ p(x)&:=x\ge x\\ q(x,y)&:=xy> xy\end{align}

The given statement is true under this interpretation (your Method 1): \begin{align}\text{universe of discourse}&:=\mathbb R\\ p(x)&:=x>x\\ q(x,y)&:=xy>xy\end{align}

Hence, the given statement is satisfiable but invalid.