1+1 = 2/Proof 2

Theorem

Define $0$ as the unique 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.

Then:

$1 + 1 = 2$

where:

$1 := \map s 0$

$2 := \map s 1 = \map s {\map s 0}$

$+$ denotes addition

$=$ denotes equality

$\map s n$ denotes the successor mapping.

Proof

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

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

By the definition of addition:

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

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

\(\text {(1)}: \quad\)

\(\ds \map s 0 + \map s 0\)

\(=\)

\(\ds \map s {\map s 0 + 0}\)

By the definition of addition:

$\forall m: m + 0 = m$

Letting $m = \map s 0$:

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

Taking the successor of both sides:

\(\text {(2)}: \quad\)

\(\ds \map s {\map s 0 + 0}\)

\(=\)

\(\ds \map s {\map s 0}\)

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

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

Hence the result.

$\blacksquare$