Prove $PA^{-}\models \forall x\forall y(y\cdot x=1\to (y\neq 1\to x=1))$

Question

Prove $PA^{-}\models \forall x\forall y(y\cdot x=1\to (y\neq 1\to x=1))$

I don't even see how this is true. Since it seems that because PA doesn't contain multiplicative inverses that if $xy=1$ then both $x=1$ and $y=1$?

Answer 1

Prove the contrapositive. In other words show that if $x\ne 1$ and $y\ne 1$ then $xy\ne 1.$ The case if $x$ or $y$ zero is easy. So assume they aren't equal to zero, so $x=u+1$ and $y=v+1$ where $u,v\ne 0.$ Then we have $$xy=(u+1)(v+1)=uv+u+v+1\ge v+1\ge 1+1>1.$$

Of course you'll need to show in some detail that all of these steps follow from the axioms. Also note that I didn't really need to assume $x$ and $y$ were both not equal to one, only that one of them was.