1+1 = 2
Jump to navigation
Jump to search
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
$1$ 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 $(1)$ | |||||||||||
$\blacksquare$
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.
$\blacksquare$