Natural Number Addition is Commutative

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


 * $\forall x, y \in \N: x + y = y + x$

Proof
Follows directly from the result that Natural Numbers under Addition is Commutative Monoid.

Alternate Proof
This result can also be proven from the Peano axioms in Peano arithmetic. To prove this, we will use mathematical induction (Axiom P5 of Peano's axioms). Addition on the natural numbers is defined recursively such that the following principles hold:


 * $\forall m \in \N ( n + 0 ) = n$
 * $\forall m, n \in \N ( n + m' ) = ( n + m )'$ where $n'$ denotes the successor of $n$.

We will prove the commutative property of addition by demonstrating that it holds for $0$, and if it holds for $n$, it holds for $n'$.

Lemma 1 - Proving the Commutative property for $0$. Starting with the fact that $( n + 0 ) = n$, we must then prove that $A = ( 0 + n )$. We will proceed by induction. $( 0 + 0 ) = 0$ by the first part of the recursive definition of addition. Suppose $( 0 + n ) = n$. Then $( 0 + n' ) = ( 0 + n )' = n'$ (which is justified by the hypothesis and substitution of equality). This completes the two parts to induction across the natural numbers.
 * $( n + 0 ) = ( 0 + n )$

Lemma 2 - Proving the Commutative property for $m'$. By the second part of the recursive definition, $( n + m' ) = ( n + m )'$. By the hypothesis, $( n + m )' = ( m + n )'$. To prove $( m + n )' = ( m' + n )$, we will, once again, employ induction. $( m + 0 )' = m' = ( m' + 0 )$, thus satisfying the case when $n = 0$. Suppose $( m + n )' = ( m' + n )$. Then $( m + n' )' = ( m + n )'' = ( m' + n )' = ( m' + n' )$.
 * $( n + m ) = ( m + n )$ implies $( n + m' ) = ( m' + n )$

Parts 1 and 2 show that the commutative property holds for 0 and all its successors, and thus the natural numbers.