1+1 = 2/Proof 2

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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

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 $n = 0$:

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


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:

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


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

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

Hence the result.

$\blacksquare$