# Addition of Natural Numbers is Provable/General Form

## Theorem

Let $y \in \N$ be a natural number.

Let $s^a$ denote the application of the successor mapping $a$ times.

Let $\sqbrk a$ denote $\map {s^a} 0$.

Then there exists a formal proof of:

$\forall x: x + \sqbrk y = \map {s^y} x$

## Proof

Proceed by induction on $y$.

### Basis for the Induction

Let $y = 0$.

Then $\sqbrk y = 0$ by definition.

Then the theorem to prove is:

$\forall x: x + 0 = x$

But that is Axiom $\text M 3$ of minimal arithmetic.

$\Box$

### Induction Hypothesis

Fix some $y \in \N$.

Suppose that there is a formal proof of:

$\forall x: x + \sqbrk y = \map {s^y} x$

### Induction Step

The following is a formal proof:

By the tableau method of natural deduction:

$\forall x: x + \map s {\sqbrk y} = \map s {\map {s^y} x}$
Line Pool Formula Rule Depends upon Notes
1 $\forall x: x + \sqbrk y = \map {s^y} x$ Theorem Introduction (None) Induction Hypothesis
2 $x_0 + \sqbrk y = \map {s^y} {x_0}$ Universal Instantiation 1
3 $\forall a,b: a + \map s b = \map s {a + b}$ Axiom $\text M 4$
4 $x_0 + \map s {\sqbrk y} = \map s {x_0 + \sqbrk y}$ Universal Instantiation 3
5 $\forall a, b: a = b \implies \map s a = \map s b$ Theorem Introduction (None) Substitution Property of Equality
6 $x_0 + \sqbrk y = \map {s^y} {x_0} \implies \map s {x_0 + \sqbrk y} = \map s {\map {s^y} {x_0} }$ Universal Instantiation 5
7 $\map s {x_0 + \sqbrk y} = \map s {\map {s^y} {x_0} }$ Modus Ponendo Ponens 6, 2
8 $x_0 + \map s {\sqbrk y} = \map s {\map {s^y} {x_0} }$ Equality is Transitive 4, 7
9 $\forall x: x_0 + \map s {\sqbrk y} = \map s {\map {s^y} x}$ Universal Generalisation 8

By expanding the definitions of $\sqbrk y$ and $\map {s^y} x$, the proven result is syntactically identical to:

$\forall x: x + \sqbrk {\map s y} = \map {s^{\map s y} } x$

which was the theorem to be proven.

$\Box$

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

$\blacksquare$