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 $(2)$

$\blacksquare$