Expressions in Peano Arithmetic that falsify these statements?

Question

I'm trying to find expressions ($P$ and $Q$) using Peano arithmetic that falsify these 'equivalences'.

I have some intuition about the nature of expressions using higher level theories, but not really in the Peano setting.

• $(P\implies(\exists!x)Q(x))\iff (\exists!x)(P\implies Q(x))$

• $(\exists!x)(\exists!y)P(x,y)\iff (\exists!y)(\exists!x)P(x,y)$

Help much appreciated.

Answer 1

For the first, let $P$ be any false statement, such as $1=2$.

For the second, how about $P(x,y)\ \equiv\ (x=y\lor x=y+1)$?

Answer 2

For the first one, I think we can try with :

$(x > 1)$ as $P$ and $(\forall y) (x \le y)$ as $Q(x)$.

We have that $(\exists ! x) (\forall y) (x \le y)$ is true, so that $(x > 1) \rightarrow (\exists ! x) (\forall y) (x \le y)$ is always true.

But $(\exists ! x)((x > 1) \rightarrow (\forall y) (x \le y))$ is false, because $(x > 1) \rightarrow (\forall y) (x \le y)$is true both for $x = 0$ (because the antecedent is false and the consequent is true) and for $x = 1$ (because both the antecedent and the consequent are false).