Axiom 2.3. \(0\) is not the sucessor of any natural number; i.e., we have \(n\pp\neq0\) for every natural number \(n\).
Axiom 2.3
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Proposition 2.2.8
\(Proof~by~Tao\). We use induction on \(b\). If \(b=0\), then \(a+b=a+0=a\), which is positive, so this proves the base case Now suppose inductively that \(a+b\) is positive. Then \(a+(b\pp)=(a+b)\pp\), which cannot be zero by Axiom 2.3, and is hence positive. This closes the induction. \(\square\)
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Exercise 2.2.3 (Propostion 2.2.12)
\(\vdash\) { Axiom 2.3: \(n\pp\neq0\) for every natural number \(n\) }