# 1+1 = 2

## Theorem

Define $0$ as the only element in the set $P \setminus s \left({P}\right)$, where:

$P$ is the Peano Structure
$s \left({P}\right)$ is the image of the mapping $s$ defined in Peano structure
$\setminus$ denotes the set difference.

The theorem to be proven is:

$1 + 1 = 2$

where:

$1 := s \left({0}\right)$
$2 := s \left({1}\right) = s \left({s \left({0}\right)}\right)$
$+$ denotes addition
$=$ denotes equality
$s \left({n}\right)$ denotes the successor function as defined by Peano

## Proof 1

$1$ is defined as $s \left({0}\right)$ and $2$ as $s \left({s \left({0}\right)}\right)$.

Therefore the statement to be proven becomes:

$s \left({0}\right) + s \left({0}\right) = s \left({s \left({0}\right)}\right)$

Thus:

 $\displaystyle \forall n \in P: \ \$ $\displaystyle m + s \left({n}\right)$ $=$ $\displaystyle s \left({m + n}\right)$ Definition of Addition $\displaystyle \implies \ \$ $\displaystyle m + s \left({0}\right)$ $=$ $\displaystyle s \left({m+0}\right)$ $\displaystyle \implies \ \$ $\displaystyle \forall m \in P: \ \$ $\displaystyle m + s \left({0}\right)$ $=$ $\displaystyle s \left({m + 0}\right)$ $\text {(1)}: \quad$ $\displaystyle \implies \ \$ $\displaystyle s \left({0}\right) + s \left({0}\right)$ $=$ $\displaystyle s \left({s \left({0}\right) + 0}\right)$ $\displaystyle \forall m \in P: \ \$ $\displaystyle m + 0$ $=$ $\displaystyle m$ Definition of Addition $\text {(2)}: \quad$ $\displaystyle \implies \ \$ $\displaystyle s \left({0}\right) + 0$ $=$ $\displaystyle s \left({0}\right)$ $\displaystyle s \left({s \left({0}\right) + 0}\right)$ $=$ $\displaystyle s \left({s \left({0}\right)}\right)$ taking the successor on both sides of $(2)$ $\text {(3)}: \quad$ $\displaystyle \implies \ \$ $\displaystyle s \left({0}\right) + s \left({0}\right)$ $=$ $\displaystyle s \left({s \left({0}\right) + 0}\right)$ from $(1)$ $\displaystyle$ $=$ $\displaystyle s \left({s \left({0}\right)}\right)$ from $(2)$

$\blacksquare$

## Proof 2

Defining $1$ as $\map s 0$ and $2$ as $\map s {\map s 0}$, the statement to be proven becomes:

$\map s 0 + \map s 0 = \map s {\map s 0}$

$\forall m \in P: \forall n \in P: m + \map s n = \map s {m + n}$

Letting $m = \map s 0$ and $n = 0$:

 $\text {(1)}: \quad$ $\displaystyle \map s 0 + \map s 0$ $=$ $\displaystyle \map s {\map s 0 + 0}$

$\forall m: m + 0 = m$

Letting $m = \map s 0$:

$\map s 0 + 0 = \map s 0$

Taking the successor of both sides:

 $\text {(2)}: \quad$ $\displaystyle \map s {\map s 0 + 0}$ $=$ $\displaystyle \map s {\map s 0}$

Applying Equality is Transitive to $(1)$ and $(2)$ we have:

$\map s 0 + \map s 0 = \map s {\map s 0}$

Hence the result.

$\blacksquare$