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:

$\forall x, y, z \in \N: x + \left({y + z}\right) = \left({x + y}\right) + z$


Proof

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

We are to show that:

$\left({x + y}\right) + n = x + \left({y + n}\right)$

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


From the definition of addition, we have that:

\(\displaystyle \forall m, n \in \N: \ \ \) \(\displaystyle m + 0\) \(=\) \(\displaystyle m\) $\quad$ $\quad$
\(\displaystyle m + n^+\) \(=\) \(\displaystyle \left({m + n}\right)^+\) $\quad$ $\quad$

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

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:

$\left({x + y}\right) + n = x + \left({y + n}\right)$


Basis for the Induction

$P \left({0}\right)$ is the case:

\(\displaystyle \left({x + y}\right) + 0\) \(=\) \(\displaystyle x + y\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x + \left({y + 0}\right)\) $\quad$ $\quad$

and so $P \left({0}\right)$ holds.

This is our basis for the induction.


Induction Hypothesis

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


So this is our induction hypothesis:

$\left({x + y}\right) + k = x + \left({y + k}\right)$


Then we need to show:

$\left({x + y}\right) + \left({k^+}\right) = x + \left({y + \left({k^+}\right)}\right)$


Induction Step

This is our induction step:


\(\displaystyle \left({x + y}\right) + k^+\) \(=\) \(\displaystyle \left({\left({x + y}\right) + k}\right)^+\) $\quad$ Definition of addition $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({x + \left({y + k}\right)}\right)^+\) $\quad$ by the induction hypothesis $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x + \left({\left({y + k}\right)^+}\right)\) $\quad$ Definition of addition $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x + \left({y + k^+}\right)\) $\quad$ Definition of addition $\quad$

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

$\blacksquare$


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