Addition of Natural Numbers is Provable

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

Then there exists formal proof of:
 * $\sqbrk x + \sqbrk y = \sqbrk {x + y}$

from the axioms of Robinson arithmetic, where $\sqbrk a$ is the unary representation of $a$.

Proof
By Unary Representation of Natural Number, let $\sqbrk a$ denote the term $\map s {\dots \map s 0}$, where there are $a$ applications of the successor mapping to the constant $0$.

Proceed by induction on $y$.

Base Case
Let $y = 0$.

Then $\sqbrk y = 0$.

The following is a formal proof:

But $x + 0 = x$.

Therefore, $\sqbrk {x + 0} = \sqbrk x$.

Thus, the proof above is a formal proof of $\sqbrk x + 0 = \sqbrk {x + 0}$.

Induction Step
Let there exist a formal proof of $\sqbrk x + \sqbrk y = \sqbrk {x + y}$.

Then, the following is a formal proof:

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

But as:
 * $\sqbrk a$

consists of $a$ applications of the successor mapping to the constant $0$, it follows that:
 * $\map s {\sqbrk y} = \sqbrk {\map s y}$

consists of $a + 1 = \map s a$ applications of $s$ to $0$.

But that is the unary representation of $\map s a$.

Therefore:
 * $\map s {\sqbrk y} = \sqbrk {\map s y}$

and
 * $\map s {\sqbrk {x + y} } = \sqbrk {\map s {x + y} }$

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

Thus:
 * $\sqbrk {\map s {x + y} } = \sqbrk {x + \map s y}$

It follows that the proof above is also a proof of:
 * $\sqbrk x + \sqbrk {\map s y} = \sqbrk {x + \map s y}$

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 $x, y \in \N$.