Understanding Invariant Simplification Steps

Question

I'm trying to make sense of the steps of a "simplification" given in a text book I'm reading.

The whole thing seems a little elaborate to explain why $p—c = (p+1) —(c+1)$. I can follow the first three steps easily enough, but the final

{ [x-y = x+(-y) ]
with x,y := p,c and with x,y := 1 , —1,
[x—x=0] with x:=1 }

leaves me confused. I can see that $1+(-1)$ is just $1-1$ but where does [x—x=0] with x:=1 come into it please?

Answer 1

The first step

$[x-y = x+(-y) ]$

with $x,y := p,c$ and with $x,y := 1 ,-1$,

implies

$p + (-c) + 1 + (-1) = p - c + 1 - 1$.

The second step does exactly what you said:

let $x$ be $1$, and from $x - x = 0$ we get $1 - 1 = 0$.