A bit of everything

# 1 plus 1 = 2

## Using Peano's axioms

By definition, we have

\begin{eqnarray} 2 &=& s(1)\\ 1 &=& s(0) \end{eqnarray}

Therefore, considering the substitution property of equality ($a = b \to f(a) = f(b)$), we have,

\begin{eqnarray} 1 + 1 = 1 + s(0) \end{eqnarray}

By Peano's axioms, we therefore have

\begin{eqnarray} 1 + 1 &=& 1 + s(0)\\ &=& s(1 + 0)\\ &=& s(1)\\ &=& 2 \end{eqnarray}

Using the transitive property of equality, this finally leads to

\begin{eqnarray} 1 + 1 = 2 \end{eqnarray}

## Using Russell's axioms

In Russell's axioms, numbers are defined as equivalence classes of sets of that cardinality. In other words, $1$ is the class of sets such that

\begin{equation} 1 = \{ A | \exists x, x \in A \wedge \forall y, z \in A, y = z \} \end{equation}

The proof used is that, given $\alpha, \beta \in 1$, we have the theorem

\begin{equation} \alpha, \beta \in 1 \to (\alpha \cap \beta = \varnothing \leftrightarrow \alpha \cup \beta \in 2) \end{equation}

## Using lambda calculus

In lambda calculus, numerals are defined by the functions which apply the next function this amount of time to the next one. In other words, $1$ is defined as

\begin{equation} 1 = \lambda f. \lambda x. f(x) \end{equation}

and $2$ is defined by

\begin{equation} 2 = \lambda f. (\lambda x. f(f(x))) \end{equation}