Natural Number Addition Commutativity with Successor/Proof 2
Theorem
Let $\N$ be the natural numbers.
Then:
- $\forall m, n \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$
Proof
Using the following axioms:
\((\text A)\) | $:$ | \(\ds \exists_1 1 \in \N_{> 0}:\) | \(\ds a \times 1 = a = 1 \times a \) | ||||||
\((\text B)\) | $:$ | \(\ds \forall a, b \in \N_{> 0}:\) | \(\ds a \times \paren {b + 1} = \paren {a \times b} + a \) | ||||||
\((\text C)\) | $:$ | \(\ds \forall a, b \in \N_{> 0}:\) | \(\ds a + \paren {b + 1} = \paren {a + b} + 1 \) | ||||||
\((\text D)\) | $:$ | \(\ds \forall a \in \N_{> 0}, a \ne 1:\) | \(\ds \exists_1 b \in \N_{> 0}: a = b + 1 \) | ||||||
\((\text E)\) | $:$ | \(\ds \forall a, b \in \N_{> 0}:\) | \(\ds \)Exactly one of these three holds:\( \) | ||||||
\(\ds a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \) | |||||||||
\((\text F)\) | $:$ | \(\ds \forall A \subseteq \N_{> 0}:\) | \(\ds \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \) |
Proof by induction:
From Axiomatization of $1$-Based Natural Numbers, we have by definition that:
\(\ds \forall m, n \in \N: \, \) | \(\ds m + 0\) | \(=\) | \(\ds m\) | |||||||||||
\(\ds \paren {m + n}^+\) | \(=\) | \(\ds m + n^+\) |
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
- $\forall m \in \N_{> 0}: \paren {m + 1} + n = \paren {m + n} + 1$
Basis for the Induction
When $n = 1$, we have:
- $\paren {m + 1} + 1 = \paren {m + 1} + 1$
which holds trivially.
Thus $\map P 1$ 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 $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis $\map P k$:
- $\forall m \in \N: \paren {m + 1} + k = \paren {m + k} + 1$
Then we need to show that $\map P {k^+}$ follows directly from $\map P k$:
- $\forall m \in \N: \paren {m + 1} + \paren {k + 1} = \paren {m + \paren {k + 1} } + 1$
Induction Step
This is our induction step:
\(\ds \paren {m + \paren {k + 1} } + 1\) | \(=\) | \(\ds \paren {\paren {m + k} + 1} + 1\) | Natural Number Addition is Associative/Proof 3 | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {m + 1} + k} + 1\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m + 1} + \paren {k + 1}\) | Natural Number Addition is Associative/Proof 3 |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall m, n \in \N_{> 0}: \paren {m + 1} + n = \paren {m + n} + 1$
$\blacksquare$