Does it translate to $$\exists x\forall y((P\rightarrow Q(x))\land ((P\rightarrow Q(y))\rightarrow y=x)),\text{ or}$$ $$\exists x\forall y(P\rightarrow (Q(x)\land (Q(y)\rightarrow y=x)))?$$

$x$ is not free in $P$.

Note: They, I’m pretty sure, are not logically equivalent. But they being not equivalent implies that $\exists !x(P\rightarrow Q(x))$ is not equivalent to $P\rightarrow \exists !xQ(x)$, which is absurd as it is allowed in prenex conversion.

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    $\begingroup$ Assuming the domain of $Q$ has more than one element, it becomes $P \wedge \exists ! Q(x)$. The failure of the conversion isn't really a surprise since $\exists !$ is a "composite". $\endgroup$ – Ian Jul 12 at 13:00
  • $\begingroup$ @Ian Can you please elaborate? Not getting it. $\endgroup$ – Atom Jul 12 at 13:03
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    $\begingroup$ Looking at this particular case, the point is that if $P$ is false then $P \rightarrow Q(x)$ is true for all $x$, which breaks the unique existence if the domain of $Q$ has more than one element. More generally the issue here is that although $\exists$ distributes over $\vee$, $\exists!$ does not. $\endgroup$ – Ian Jul 12 at 19:16

I would say the meaning is the first one.

What about your assertion

$\exists !x(P\rightarrow Q(x))$ is not equivalent to $P\rightarrow \exists !xQ(x)$

What if $P$ is false and $Q(x)$ is "$x = 0$". [Certainly $\exists ! x Q(x)$ is true. But so what?] Then $P \rightarrow S$ is true for any $S$. In particular $P \rightarrow Q(x)$ is true for all $x$, and thus $\exists !x(P\rightarrow Q(x))$ is false because uniqueness fails. On the other hand, as noted, $P \rightarrow S$ is true for any $S$, so in particular $P\rightarrow \exists !xQ(x)$ is true.

So you are correct, they are not equivalent. So I guess your next assertion

which is absurd

is wrong.


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