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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 :)

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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:

$x+1=1+x$.

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

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