Define the following predicate P(x,y,z) for naturals x and y and positive naturals z.
If x < z, P(x,y,z) is true $iff$ x = y.
If y > = z, then P(x,y,z) is false.
If y < z and x >= z, then P(x,y,z) is true $iff$ P(x - z,y,z) is true.
Prove that if y < z, then P(x,y,z) is true $iff$ $(\exists r: x = rz+ y$) where r is $n\in\mathbb{N}$
What is the best way to approach this?
The best approach is an if-then argument in both directions:
First show that if $P(x,y,z)$ is true, there exists $r$ such that $x=rz+y$.
Then show that if there exists such an $r$, $P(x,y,z)$ is true.
Or maybe in the opposite order.
For instance, what is $P(rz+y,y,z)$? By definition of $P$, this is true if and only if $P(rz+y-z,y,z)$. Keep going.