Axiom 2.4. Different natural numbers must have different successors; i.e., if \(n,~m\) are natural numbers and \(n\neq m\), then \(n\pp\neq m\pp\). Equivalently, if \(n\pp=n\pp\), then we must have \(n=m\).
Axiom 2.4
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Proposition 2.2.6
\(Proof~by~Tao\). We prove this by induction on \(a\). First consider the base case \(a=0\). Then we have \(0+b=0+c\), which by definition of addition implies that \(b=c\) as desired. Now suppose inductively that we have the cancellation law for \(a\) (so that \(a+b=a+c\) implies \(b=c\)); we now have to prove the cancellation law for \(a\pp\). In other words, we assume that \((a\pp)+b=(a\pp)+c\) and need to show that \(b=c\). By the definition of addition, \((a\pp)+b=(a+b)\pp\) and \((a\pp)+c=(a+c)\pp\) and so we have \((a+b)\pp=(a+c)\pp\). By Axiom 2.4, we have \(a+b=a+c\). Since we already have the cancellation law for \(a\), we thus have \(b=c\) as desired. This closes the induction. \(\square\)
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Exercise 2.2.2 (Lemma 2.2.10)
\(\Rightarrow\) {\(b=a\), \(c=a\) by Axiom 2.4 }