1+1 = 2/Proof 1

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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

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

Hence the statement to be proven becomes:

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

Thus:

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

$\blacksquare$