Prove that order is antisymmetric. (for natural numbers) Prove that order is antisymmetric.(for natural numbers)i.e.
If $ a \leq b$ and $b\leq a$ then $a=b$.
I do not want a proof based on set theory.
I am following the book Analysis 1 by Tao. It should be based on peano axioms.
I tried $ b=a+n$ where $n$ is a natural number then $ a+n \leq a$ but subtraction not yet defined (in the text that I am following).
How should I proceed ?

If you're allowed to rely on trichotomy you might as well suppose by way of contradiction that $a\ne b$. Since $a\leq b$ and $b\leq a$ it follows that $a<b$ and $b<a$, a contradiction to the forementioned trichotomy property.
Remarks.

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*For this proof to be valid you'd sooner have $a\leq b$ defined as $a<b\oplus a=b$ where by $\oplus$ we mean an exclusive or

*To understand why we can conclude $a<b$ and $b<a$ from $a\leq b$ and $b\leq a$ take into account that $(p\oplus q)\wedge \sim q\implies p$ where $p$ and $q$ are propositions and $\oplus$ is an exclusive or.

