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:
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 $y \in \N$.
$\blacksquare$