Proof by induction, the difference of the same number is zero

Question

So I need to proof $n - n = 0$ for all n in the natural numbers, what we are allowed to use is: The predecessor and the $(-)$ operation

The predecessor base case: $P(0) = 0$

The predecessor step case: $P(S(n)) = n$, where $S(n)$ is the successor of $n$

$(-)$ base case: $n - 0 = n$

$(-)$ step case: $n - S(m) = P(n-m)$

My attempt:

For the base case: $0-0=0$ using base case of $(-)$

My induction hypothesis is that $n-n=0$ for all n in the natural numbers

Then my step case: $$S(n)-S(n)=P(S(n) -n)$$
I'm stuck right here I didn't know what to apply next. If I was able to switch $S(n)-n$ then I'll be able to apply the step case of $(-)$ then would have got my answer

Answer 1

What we want to use is that $P(j-k) = P(j)-k$. This would be useful in your last step as then $S(n)-S(n) = P(S(n) - n) = P(S(n))-n = n-n$. So we only have to prove that which is done by induction over $k$:

First we have that $P(j-0) = P(j) = P(j)-0$ by the definitions. Now assume that $P(j-k) = P(j)-k$ then we have:

$$P(j-S(k)) = P(P(j-k)) = P(P(j)-k)) = P(j)-S(k)$$

which proves the statement is true for the successor of $k$ as well and therefore by induction for all $k$.

To complete your proof we have that $S(n)-S(n) = P(S(n)-n) = P(S(n)) - n = n - n = 0$ which means that if the statement is true for $n$ then it's true for $S(n)$ too and by induction true for all $n$.