Prove that it is a theorem

Question

I'm doing some excercises from the book "The Incompleteness Phenomenom" from Goldstern and Judah.

I'm doing exercise about Peano Arithmetic and I have to show that:

1. $\vdash p|x \wedge r·s=p \to r|x$
2. $\vdash r·s=p \wedge r\geq p > 1\to s=1$

I'm very obfuscated with derivations and I don't know how to start T.T

Answer 1

I hope you have all simple but useful results available that are used below:

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From $p\mid x$, $\exists z\,p\cdot z=x$. For this $z$, $p\cdot z=x$ and $r\cdot s=p$ imply $(r\cdot s)\cdot z=x$ and by associativity of multiplication $r\cdot(s\cdot z)=x$, hence there exists $z'$ (namely $z=sz$) such that $r\cdot z'=x$, i.e., $r\mid x$.

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If $s=0$ then $p=r\cdot s=0$, but $p>1$. Hence $s\ge 1$ and we can write $s=1+t$. Then $r\ge p=r\cdot s=r\cdot (1+t)=r+r\cdot t\ge r$, hence $r+r\cdot t=r$ and by cancellation $r\cdot t=0$. Then $r=0$ or $t=0$ and so (as $r\ge p>1$) $t=0$, i.e., $s=1$.