Natural Number Addition Commutativity with Successor

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Theorem

Let $\N$ be the natural numbers.

Then:

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


Proof 1

Proof by induction:


From the definition of addition, we have that:

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


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

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


Basis for the Induction

By definition, we have:

\(\displaystyle \forall m \in \N: \ \ \) \(\displaystyle m^+ + 0\) \(=\) \(\displaystyle m^+\) by definition
\(\displaystyle \) \(=\) \(\displaystyle \left({m + 0}\right)^+\) by definition: $m = m + 0$

Thus $P \left({0}\right)$ is seen to be true.


This is our basis for the induction.


Induction Hypothesis

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


So this is our induction hypothesis $P \left({k}\right)$:

$\forall m \in \N: m^+ + k = \left({m + k}\right)^+$


Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$:

$\forall m \in \N: m^+ + k^+ = \left({m + k^+}\right)^+$


Induction Step

This is our induction step:


\(\displaystyle m^+ + k^+\) \(=\) \(\displaystyle \left({m^+ + k}\right)^+\) by definition
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({m + k}\right)^+}\right)^+\) from the induction hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \left({m + k^+}\right)^+\) by definition

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


Therefore:

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

$\blacksquare$


Proof 2

Using the following axioms:

\((A)\)   $:$     \(\displaystyle \exists_1 1 \in \N_{> 0}:\) \(\displaystyle a \times 1 = a = 1 \times a \)             
\((B)\)   $:$     \(\displaystyle \forall a, b \in \N_{> 0}:\) \(\displaystyle a \times \paren {b + 1} = \paren {a \times b} + a \)             
\((C)\)   $:$     \(\displaystyle \forall a, b \in \N_{> 0}:\) \(\displaystyle a + \paren {b + 1} = \paren {a + b} + 1 \)             
\((D)\)   $:$     \(\displaystyle \forall a \in \N_{> 0}, a \ne 1:\) \(\displaystyle \exists_1 b \in \N_{> 0}: a = b + 1 \)             
\((E)\)   $:$     \(\displaystyle \forall a, b \in \N_{> 0}:\) \(\displaystyle \)Exactly one of these three holds:\( \)             
\(\displaystyle a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \)             
\((F)\)   $:$     \(\displaystyle \forall A \subseteq \N_{> 0}:\) \(\displaystyle \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \)             


Proof by induction:


From Axiom:Axiomatization of 1-Based Natural Numbers, we have by definition that:

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


For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:

$\forall m \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$


Basis for the Induction

When $n = 1$, we have:

$\left({m + 1}\right) + 1 = \left({m + 1}\right) + 1$

which holds trivially.

Thus $P \left({1}\right)$ is seen to be true.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.


So this is our induction hypothesis $P \left({k}\right)$:

$\forall m \in \N: \left({m + 1}\right) + k = \left({m + k}\right) + 1$


Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$:

$\forall m \in \N: \left({m + 1}\right) + \left({k + 1}\right) = \left({m + \left({k + 1}\right)}\right) + 1$


Induction Step

This is our induction step:


\(\displaystyle \left({m + \left({k + 1}\right)}\right) + 1\) \(=\) \(\displaystyle \left({\left({m + k}\right) + 1}\right) + 1\) Natural Number Addition is Associative/Proof 3
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({m + 1}\right) + k}\right) + 1\) from the induction hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \left({m + 1}\right) + \left({k + 1}\right)\) Natural Number Addition is Associative/Proof 3

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall m, n \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$

$\blacksquare$


Also defined as

Some treatments of Peano's axioms define the non-successor element (or primal element) to be $1$ and not $0$.

The treatments are similar, but the $1$-based system results in an algebraic structure which has no identity element for addition, and so no zero for multiplication.


Thus, in the context of 1-based natural numbers, this result can be written:

$\forall m, n \in \N_{> 0}: \paren {m + 1} + n = \paren {m + n} + 1$