# What does $\exists !x(P\rightarrow Q(x))$ translate to? A paradox?

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.

• 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". – Ian Jul 12 at 13:00
• @Ian Can you please elaborate? Not getting it. – Atom Jul 12 at 13:03
• 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. – Ian Jul 12 at 19:16

I would say the meaning is the first one.

$$\exists !x(P\rightarrow Q(x))$$ 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.