# prove $x+1 = 1+x$ using Peano axioms

I wanna prove that $$x+1 = 1+x$$ (without considering "$$x+0=x$$",and Im using the old definition of Peano axioms)
This is my try:
Using this basis:

$$(1):1+x = x^+$$
$$(2):x^+ +y=(x+y)^+$$

Actually my idea is if numbers succesor are equal then actual numbers are equal too. (based on Peano axioms) $$(1+x)^+ = 1^+ + x = (1+1)+x$$
and $$(x+1)^+ = x^+ +1 = (1+x)+1$$
But I stucked in how to show that this two are equal.

Somebody help :)

We will prove by induction that $$x+1=x^+$$.

Basis: $$1+1=1^+$$, from (1).

Induction step: assume $$x+1=x^+$$ and prove that $$x^++1=(x^+)^+$$.

By (2): $$x^++1 = (x+1)^+ = (x^+)^+$$ using induction hypotheses and substitution for equality.

Having proved that $$x+1=x^+$$, for every $$x$$, we use (1) and transitivity and symmetry of equality to conclude that:

$$x+1=1+x$$.

• Thanks!Clear and easy to understand. – program_craft Feb 25 at 16:18
• @program_craft - you are welcome :-) – Mauro ALLEGRANZA Feb 25 at 16:19