Multiplication of Natural Numbers is Provable

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

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

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 $x \times 0 = 0$.

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

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

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

Then the following is a formal proof:

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

But $\sqbrk y$ consists of $y$ applications of the successor mapping to the constant $0$.

Thus, $\map s {\sqbrk y}$ consists of $y + 1 = \map s y$ mappings of $s$ to $0$.

As that is the unary representation of $\map s y$ it follows that:
 * $\map s {\sqbrk y} = \sqbrk {\map s y}$

By definition of multiplication:
 * $x \times \map s y = \paren {x \times y} + x$

Therefore, the above proof also proves:
 * $\sqbrk x \times \sqbrk {\map s y} = \sqbrk {x \times \map s y}$

as the WFFs are syntactically identical.

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 $x, y \in \N$.