Operator precedence - Discrete Math (Predicate logic)

Question

Can someone please give me a hint as to how these two statements are different? Thank you!

∀x ∈ S, ∃y ∈ T, P(x, y) ⇒ Q(x) (Statement 1)

∀x ∈ S, (∃y ∈ T, P(x, y)) ⇒ Q(x) (Statement 2)

Answer 1

The first one implies that for all $x$ in $S$ there exists a $y$ in $T$ such that the predicate $P(x,y)$ implies $Q(x)$ whereas in the second one, the existence of a $y$ in $T$ such that $P(x,y)$ is true implies $Q(x)$. The first one is a consequence of $P(x)$ being true, the other one is a consquence of the existence of an element that makes the predicate true.

Answer 2

Let's consider a concrete case $S=T=\mathbb{R}$, $P(x,y)$ is $xy=0$ and $Q(x)$ is $x=0$.

Then statement 1 says: for any number $x$, I can find a test number $y$ such that if $xy=0$ then actually $x=0$. This is true: take $y=1$.

Statement 2 says: for any number $x$, if I can find a number $y$ such that $xy=0$, then actually $x$ must be $0$. This is false, since for any $x$ I can always take $y=0$, and then $xy=0$ but $x$ may be nonzero.

Answer 3

Statement 1 is $\forall x{\in}S~\Big(\forall y{\in}T~\color{red}{\big(}P(x,y)\to Q(x)\color{red}{\big)}\Big)$ and the conditional is inside the scope of the second universal quantifier.

Statement 2 is $\forall x{\in}S~\Big(\color{red}{\big(}\forall y{\in} T~P(x,y)\color{red}{\big)}\to Q(x)\Big)$ and here the second universal quantifier is inside the antecedent of the conditional.

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Now, recall that a conditional is satisfied if the consequent is satisfied or the antecedent is unsatisfied.

$\forall y~(P(x,y)\to Q(x))$ is satisfied if $Q(x)$ is satisfied or $P(x,y)$ is unsatisfied by every $y$.

$(\forall y~P(x,y))\to Q(x)$ is satisfied if $Q(x)$ is satisfied or $P(x,y)$ is unsatisfied by some $y$.