Are these expressions logically equivalent?

Question

The statement goes, "Let $H \leq G$ be groups such that $H \setminus \{1\}$ does not intersect any conjugate of (some subset of $G$) $T$. Then [conclusion about $H$]."

I want to formulate this "if" condition logically. First I wrote it as: $$ \forall h \in H: \left[\,\,\left(\exists g \in G: g^{-1}hg \in T \right) \implies h = 1\right]. $$

But then I thought it was wrong and I instead put $$ \forall h \in H \,\,\,\,\forall g \in G: \left[g^{-1}hg \in T \implies h = 1\right]. $$

Now I'm confused and I don't know which is correct, or if they are equivalent (or one is stronger than the other). The first expression seemed more intuitive ("if an element of $H$ can be conjugated inside $T$, it must be $1$), but the second expression seems to follow the statement more ("For any $g$, the conjugation $g^{-1}Hg$ only intersects $T$ in the unit").

Thank you!

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EDIT: If this helps, this could be abstracted to the following: is the expression $$ \forall x [[\exists y: P(x,y)] \implies Q(x)] $$ logically equivalent to $$ \forall x \forall y [P(x,y) \implies Q(x)]? $$

Answer 1

Let's fix $x$.

Then both sentences $[[\exists y: P(x,y)] \implies Q(x)]$ and $\forall y [P(x,y) \implies Q(x)]$ may be rewritten as $Q(x)\vee \neg(\exists y : P(x,y))$.

The first one is trivial by definition of $\implies$. The second one is not hard too : "either $Q(x)$ is true, either not. What would happen if $Q(x)$ was not true ?".

Thus for all $x$, the subsentences are equivalent so the answer, for me, is yes.

Answer 2

Here's another way to think about it that doesn't require rewriting $A\to B$ as $\lnot A\lor B$.

It suffices to show that $(\exists y\, P(x,y))\to Q(x)$ is equivalent to $\forall y\,(P(x,y)\to Q(x))$.

Suppose $(\exists y\, P(x,y))\to Q(x)$. Fix some $b$ and assume $P(x,b)$. Then $\exists y\, P(x,y)$ is true, witnessed by $b$, so $Q(x)$ is true. Since $b$ was arbitrary, $\forall y\, (P(x,y)\to Q(x))$.

Conversely, suppose $\forall y\, (P(x,y)\to Q(x))$. Assume $\exists y \,P(x,y)$. Let $b$ be a witness, so $P(x,b)$ is true. Then since $\forall y\, (P(x,y)\to Q(x))$, in particular $P(x,b)\to Q(x)$, so $Q(x)$ is true. Thus $(\exists y\, P(x,y))\to Q(x)$.