How do you prove this logical equivalence?

Question

$\\ (\exists! x:P(x)) \leftrightarrow ((\forall x:P(x) \rightarrow Q(x))\leftrightarrow(\exists x:P(x) \land Q(x)))$

If there's only one $x$ for which $P(x)$, then saying "all $x$ for which $P(x)$, $Q(x)$" is the same as saying "there's an $x$ for which both $P(x)$ and $Q(x)$"

Edit: The first $\leftrightarrow$ was actually meant to be $\rightarrow$

Answer 1

Regarding your last comment:

1) $∃!xP(x)$ --- premise

2) $∃x[P(x) \land \forall y (P(y) \to x=y)]$ --- from 1)

3) $∀x (P(x)→Q(x))$ --- assumed [a]

4) $P(x) \land \forall y (P(y) \to y=x)$ --- assumed [b] from 2) for $\exists$-elim

5) $P(x)→Q(x)$ --- from 4) by $\forall$-elim

6) $P(x)$ --- from 4) by $\land$-elim

7) $Q(x)$ --- from 5) and 6) by $\to$-elim

8) $P(x) \land Q(x)$ --- from 6) and 7) by $\land$-intro

9) $\exists x (P(x) \land Q(x))$ --- from 8) by $\exists$-intro

10) $\exists x (P(x) \land Q(x))$ --- from 2), 4) and 9) by $\exists$-elim, discharging [b]

$∀x (P(x)→Q(x)) \to \exists x (P(x) \land Q(x))$ --- from 4) and 10) by $\to$-intro, discharging [a].

Conclusion:

$∃!xP(x) \vdash ∀x (P(x)→Q(x)) \to \exists x (P(x) \land Q(x))$.

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For the other part needed for the bi-conditional, we can prove it by contradiction:

1) and 2) as above

3) $\exists x (P(x) \land Q(x))$ --- assumed [a]

4) $\lnot ∀x (P(x) → Q(x))$ --- assumed [b]

5) $\exists x (P(x) \land \lnot Q(x))$ --- from 4)

6) $P(a) \land \forall y (P(y) \to a=y)$ --- assumed [c] from 2) for $\exists$-elim

Now we need two new "nested" $\exists$-elims from 3) and 5), deriving respectively $P(b) \land Q(b)$ and $P(c) \land \lnot Q(c)$, with $a,b,c$ all distinct.

From $P(b)$ and $\forall y(P(y) \to a=y)$ from 2) we derive: $a=b$.

From $P(c)$ and $\forall y(P(y) \to a=y)$ from 2) we derive: $a=c$.

From $a=b$ and $Q(b)$ by $=$-elim we derive $Q(a)$.

From $a=c$ and $\lnot Q(c)$ by $=$-elim we derive $\lnot Q(a)$.

Thus, we have $Q(a)$ and $\lnot Q(a)$ and we derive: $\bot$ by $\to$-elim [$\lnot \phi$ is equivalent to $\phi \to \bot$]

In $\bot$ there are no occurrences of $a,b,c$; thus we can correctly conclude from 3), 5) and 6) by $\exists$-elim thrice with:

7) $\bot$

8) $∀x (P(x) → Q(x))$ --- from 4) and 7) by Double Negation, discharging [b]

$\exists x (P(x) \land Q(x)) \to ∀x (P(x) → Q(x))$ --- from 3) and 8) by $\to$-intro, discharging [a].

Conclusion:

$∃!xP(x) \vdash \exists x (P(x) \land Q(x)) \to ∀x (P(x) → Q(x))$.

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Thus, we have proved:

$∃!xP(x) \vdash ∀x (P(x) → Q(x)) \leftrightarrow \exists x (P(x) \land Q(x))$.

One final application of $\to$-intro and we have proved that:

if there's only one $x$ which is $P$, then to assert that "all $x$ which are $P$ are $Q$" is the same as "there's an $x$ which is both $P$ and $Q$".

Answer 2

$$(\exists! x~:~P(x)) \iff (\forall x~:~P(x) \implies Q(x)) \iff (\exists x~:~P(x) \land Q(x))$$

The first thing to check is if it is actually true. $\iff$ is associative and commutative, it just means that an even number of statements are false.

$P$ and $Q$ are unary predicates, so they are interchangable with unary sets (i.e., $P(x)$ is the same as $x \in P$). So $\exists! x ~:~ P(x)$ is the same as $|P| = 1$, and $\forall x~:~P(x) \implies Q(x)$ is the same $P \subseteq Q$, and $\exists x~:~P(x) \land Q(x)$ is the same as $P \cap Q \ne \emptyset$.

So the claim is equivalent to the claim that an odd number of the following statements are true:

• $|P| = 1$
• $P \subseteq Q$
• $P \cap Q \ne \emptyset$

So just trial and error, eventually trying out the first statement to be false and the last two true:

$$P = \{2, 3\}$$ $$Q = P$$

Is a counter example to the proposition.