1+1 = 2/Proof 1
< 1+1 = 2
Jump to navigation
Jump to search
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
$1$ is defined as $s \left({0}\right)$ and $2$ as $s \left({s \left({0}\right)}\right)$.
Therefore the statement to be proven becomes:
- $s \left({0}\right) + s \left({0}\right) = s \left({s \left({0}\right)}\right)$
Thus:
| \(\ds \forall n \in P: \ \ \) | \(\ds m + s \left({n}\right)\) | \(=\) | \(\ds s \left({m + n}\right)\) | Definition of Addition | ||||||||||
| \(\ds \implies \ \ \) | \(\ds m + s \left({0}\right)\) | \(=\) | \(\ds s \left({m+0}\right)\) | |||||||||||
| \(\ds \implies \ \ \) | \(\ds \forall m \in P: \ \ \) | \(\ds m + s \left({0}\right)\) | \(=\) | \(\ds s \left({m + 0}\right)\) | ||||||||||
| \(\text {(1)}: \quad\) | \(\ds \implies \ \ \) | \(\ds s \left({0}\right) + s \left({0}\right)\) | \(=\) | \(\ds s \left({s \left({0}\right) + 0}\right)\) | ||||||||||
| \(\ds \forall m \in P: \ \ \) | \(\ds m + 0\) | \(=\) | \(\ds m\) | Definition of Addition | ||||||||||
| \(\text {(2)}: \quad\) | \(\ds \implies \ \ \) | \(\ds s \left({0}\right) + 0\) | \(=\) | \(\ds s \left({0}\right)\) | ||||||||||
| \(\ds s \left({s \left({0}\right) + 0}\right)\) | \(=\) | \(\ds s \left({s \left({0}\right)}\right)\) | taking the successor on both sides of $(2)$ | |||||||||||
| \(\text {(3)}: \quad\) | \(\ds \implies \ \ \) | \(\ds s \left({0}\right) + s \left({0}\right)\) | \(=\) | \(\ds s \left({s \left({0}\right) + 0}\right)\) | from $(1)$ | |||||||||
| \(\ds \) | \(=\) | \(\ds s \left({s \left({0}\right)}\right)\) | from $(2)$ |
$\blacksquare$