Definition 2.2.1. (Addition of natural numbers). Let \(m\) be a natural number. To add zero to \(m\), we define \(0+m:=m\). Now suppose inductively that we have defined how to add \(n\) to \(m\). Then we can add \(n\pp\) to \(m\) by defining \((n\pp)+m:=(n+m)\pp\)
Definition 2.2.1
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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 2.3.5 (Proposition 2.3.9)
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Exercise 2.2.4 (Propostion 2.2.13)
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Exercise 2.2.3 (Propostion 2.2.12)
\(\equiv\) { definition of addition: \((n\pp)+m := (n+m)\pp\) }
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Exercise 2.2.1 (Proposition 2.2.5)
\(=\) { definition of addition: \((n\pp)+m := (n+m)\pp\) }
\(\equiv\) { definition of addition: \(0 + m := m\) }