# 1+1 = 2/Proof 1

## Theorem

Define $0$ as the only element in the set $P \setminus \map s P$, where:

$P$ is the Peano Structure
$\map s P$ 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:

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

## Proof

$1$ is defined as $\map s 0$ and $2$ as $\map s {\map s 0}$.

Therefore the statement to be proven becomes:

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

Thus:

 $\ds \forall n \in P: \,$ $\ds m + \map s n$ $=$ $\ds \map s {m + n}$ Definition of Addition in Peano Structure $\ds \leadsto \ \$ $\ds m + \map s 0$ $=$ $\ds \map s {m + 0}$ $\ds \leadsto \ \$ $\ds \forall m \in P: \,$ $\ds m + \map s 0$ $=$ $\ds \map s {m + 0}$ $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds \map s 0 + \map s 0$ $=$ $\ds \map s {\map s 0 + 0}$ $\ds \forall m \in P: \,$ $\ds m + 0$ $=$ $\ds m$ Definition of Addition in Peano Structure $\text {(2)}: \quad$ $\ds \leadsto \ \$ $\ds \map s 0 + 0$ $=$ $\ds \map s 0$ $\ds \map s {\map s 0 + 0}$ $=$ $\ds \map s {\map s 0}$ taking the successor on both sides of $(2)$ $\text {(3)}: \quad$ $\ds \leadsto \ \$ $\ds \map s 0 + \map s 0$ $=$ $\ds \map s {\map s 0 + 0}$ from $(1)$ $\ds$ $=$ $\ds \map s {\map s 0}$ from $(2)$

$\blacksquare$