1+1 = 2

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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

$1$ 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 $(2)$

$\blacksquare$


Proof 2

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

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


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.

$\blacksquare$