Corollary 2.2.9 If \(a\) and \(b\) are natural numbers such that \(a+b=0\), then \(a=0\) and \(b=0\).
\(Proof~by~Tao\). Suppose for the sake of contradiction that \(a\neq0\) or \(b\neq0\). If \(a\neq0\) then \(a\) is positive, and hence \(a+b\) is positive by Propostion 2.2.8, a contradiction. Similarly if \(b\neq0\) then \(b\) is positive, and again \(a+b\) is positive by Proposition 2.2.8, a contradiction. Thus \(a\) and \(b\) must both be zero. \(\square\)
\(Proof~in~structured~derivation\)
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\(\bullet\) Show that \(a,b\in\mathbb{N}.~a+b=0\implies a=0\land b=0\).
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\(\Vdash\) { Proof by contradiction }
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\(\bullet\) Show that there is a contradiction, when
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– \(a+b=0\)
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– \(a\neq0\lor b\neq0\)
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\(\Vdash\) { Proof by cases }
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\(\bullet\) Show that there is a contradiction, when
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– \(a+b=0\)
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– \(a\neq0\)
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\(\Vdash\) \(a+b\)
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\(\neq\) { Proposition 2.2.8: \(a,b\in\mathbb{N}.~a\neq0\implies a+b\neq0\) }
- 0
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\(\vdash\) { assumption: \(a+b=0\) }
- \(\bot\)
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\(\square\)
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\(\bullet\) Show that there is a contradiction, when
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– \(a+b=0\)
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– \(b\neq0\)
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\(\Vdash\) \(a+b\)
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\(\neq\) { Proposition 2.2.8: \(a,b\in\mathbb{N}.~b\neq0\implies a+b=b+a\neq0\) }
- 0
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\(\vdash\) { assumption \(a+b=0\) }
- \(\bot\)
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\(\square\)
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\(\square\)
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\(\square\)