Natural Number Addition Commutativity with Successor/Proof 1

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Theorem

Let $\N$ be the natural numbers.

Then:

$\forall m, n \in \N: m^+ + n = \paren {m + n}^+$


Proof

Proof by induction:


From definition of addition:

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


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

$\forall m \in \N: m^+ + n = \paren {m + n}^+$


Basis for the Induction

From definition of addition:

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

Thus $\map P 0$ is seen to be true.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.


So this is our induction hypothesis $\map P k$:

$\forall m \in \N: m^+ + k = \paren {m + k}^+$


Then we need to show that $\map P {k^+}$ follows directly from $\map P k$:

$\forall m \in \N: m^+ + k^+ = \paren {m + k^+}^+$


Induction Step

This is our induction step:


\(\ds m^+ + k^+\) \(=\) \(\ds \paren {m^+ + k}^+\) Definition of Addition
\(\ds \) \(=\) \(\ds \paren {\paren {m + k}^+}^+\) induction hypothesis
\(\ds \) \(=\) \(\ds \paren {m + k^+}^+\) Definition of Addition

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


Therefore:

$\forall m, n \in \N: m^+ + n = \paren {m + n}^+$

$\blacksquare$


Sources

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