Addition of Natural Numbers is Provable

Theorem
Let $x, y \in \N$ be natural numbers.

Let $z = x + y$.

Then there exists formal proof of $x + y = z$ from the axioms of Robinson arithmetic.

Proof
By Unary Representation of Natural Number, assume that each $x, y, z$ consists of finitely many applications of the successor mapping to the constant $0$.

Proceed by induction on $y$.

Base Case
Let $y = 0$.

Then $z = x + 0 = x$ by the definition of addition.

But then $x$ and $z$ are precisely the same term, as described above, by Leibniz's Law.

Therefore, a formal proof that $x + 0 = x$ is identical to a formal proof that $x + 0 = z$.

The following is such a formal proof:

Therefore, there exists a formal proof of $x + 0 = z$.

Induction Step
Let there exist a formal proof of $x + y = z$.

By the definition of addition, $x + \map s y = \map s {x + y} = \map s z$.

Then, the following is a formal proof:

Therefore, there exists a formal proof of $x + \map s y = \map s z$.

Thus, by the Principle of Mathematical Induction, there exists such a formal proof for every $y \in \N$.

As $x$ was arbitrary, there exists such a formal proof for every $x, y \in \N$