2.2.4. Justify the three statments marked (why?) in the proof of Proposition 2.2.13.
\(0\leq b\) for all \(b\).
- \(\bullet\) Show that \(0\leq b\) for all \(b\)
- \(\Vdash\) \(b\)
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\(=\) { definition of addition: \(0+m:=m\) }
- \(0+b\)
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\(\Rightarrow\) { definition of ordering: \(m\leq n\) iff \(n=m+a\) for some natural number \(a\) }
- \(b\geq 0\)
- \(\square\)
If \(a>b\), then \(a\pp>b\).
-
\(\bullet\) Show that \(a>b\implies a\pp>b\)
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\(\Vdash\) \(a>b\)
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\(\equiv\) { definition of ordering: \(m<n\) iff \((m\leq n)\land(m\neq n)\) }
- \((a\geq b)\land(a\neq b)\)
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\(\equiv\) { properties of order (d): \(a\geq b\) iff \(a+c\geq b+c\) }
- \((a+1\geq b+1)\land(a\neq b)\)
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\(\equiv\) { \(n\pp=n+1\) }
- \((a\pp\geq b\pp)\land(a\neq b)\)
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\(\equiv\) { properties of order (e): \(a<b\) iff \(a\pp\leq b\) }
- \((a\pp>b)\land(a\neq b)\)
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\(\Rightarrow\) { \(P\land Q\implies P\) }
- \(a\pp>b\)
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\(\square\)
If \(a=b\), then \(a\pp>b\).
- \(\bullet\) Show that \(a=b\implies a\pp>b\)
- \(\Vdash\) \(a=b\)
-
\(\Rightarrow\) { Substitution axiom: \(a=b\implies f(a)=f(b)\) }
- \(a\pp=b\pp\)
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\(\equiv\) { \(n\pp=n+1\) }
- \(a\pp=b+1\)
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\(\equiv\) { properties of order (f): \(a<b\) iff \(b=a+d\) for some positive number \(d\) }
- \(a\pp>b\)
- \(\square\)