Natural Number Addition is Associative/Proof 2

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Theorem

The operation of addition on the set of natural numbers $\N$ is associative:

$\forall x, y, z \in \N: x + \paren {y + z} = \paren {x + y} + z$


Proof

Consider the von Neumann construction of natural numbers $\N$, as elements of the minimal infinite successor set $\omega$.

We are to show that:

$\paren {x + y} + n = x + \paren {y + n}$

for all $x, y, n \in \N$.


From the definition of addition, we have that:

\(\ds \forall m, n \in \N: \ \ \) \(\ds m + 0\) \(=\) \(\ds m\)
\(\ds m + n^+\) \(=\) \(\ds \paren {m + n}^+\)

Let $x, y \in \N$ be arbitrary.

For all $n \in \N$, let $\map P n$ be the proposition:

$\paren {x + y} + n = x + \paren {y + n}$


Basis for the Induction

$\map P 0$ is the case:

\(\ds \paren {x + y} + 0\) \(=\) \(\ds x + y\)
\(\ds \) \(=\) \(\ds x + \paren {y + 0}\)

and so $\map P 0$ holds.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, then it logically follows that $\map P {k^+}$ is true.


So this is our induction hypothesis:

$\paren {x + y} + k = x + \paren {y + k}$


Then we need to show:

$\paren {x + y} + \paren {k^+} = x + \paren {y + \paren {k^+} }$


Induction Step

This is our induction step:


\(\ds \paren {x + y} + k^+\) \(=\) \(\ds \paren {\paren {x + y} + k}^+\) Definition of Addition in Minimal Infinite Successor Set
\(\ds \) \(=\) \(\ds \paren {x + \paren {y + k} }^+\) induction Hypothesis
\(\ds \) \(=\) \(\ds x + \paren {\paren {y + k}^+}\) Definition of Addition in Minimal Infinite Successor Set
\(\ds \) \(=\) \(\ds x + \paren {y + k^+}\) Definition of Addition in Minimal Infinite Successor Set

So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction.

$\blacksquare$


Sources