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Proofs by contradiction

A common indirect technique for proving a statement \bgroup\color{red}$F$\egroup is to reason by contradiction: we assume that \bgroup\color{red}$F$\egroup does not hold and proceed to show that this is an absurd hypothesis, in that it allows us to derive a contradiction. NDL models such reasoning with proofs of the form suppose-absurd \bgroup\color{red}$F$\egroup \bgroup\color{red}$D$\egroup. To evaluate such a proof in an assumption base \bgroup\color{red}$\beta$\egroup, we add the hypothesis \bgroup\color{red}$F$\egroup to \bgroup\color{red}$\beta$\egroup and proceed to evaluate the body \bgroup\color{red}$D$\egroup in \bgroup\color{red}$\mbox{$\beta$}\cup \{F\}$\egroup. If that produces the contradiction false, then we return \bgroup\color{red}$\mbox{\tt\symbol{126}}F$\egroup as the conclusion:
\bgroup\color{red}$\;\;$\egroup \bgroup\color{red}$\mbox{$\beta$}\cup \{F\} \mbox{$\:\vdash\:$}D \mbox{$\:\longrightarrow\;$}\mbox{\tt false}$\egroup \bgroup\color{red}$\;$\egroup
\bgroup\color{red}$\;\:$\egroup
\bgroup\color{red}$\;\;$\egroup \bgroup\color{red}$\mbox{$\beta$}\mbox{$\:\vdash\:$}\mbox{\tt suppose-absurd } \, F \;\, D \mbox{$\:\longrightarrow\;$}\mbox{\tt\symbol{126}}F$\egroup \bgroup\color{red}$\;$\egroup
Note that false can be derived if the assumption base contains two formulas of the form \bgroup\color{red}$F$\egroup and \bgroup\color{red}$\mbox{\tt\symbol{126}}F$\egroup: we can then apply the rule absurd to these two formulas and derive false (recall the semantics of absurd from Section 1.10.2). As a simple example, here is a proof of A ==>   A:
assume A
  suppose-absurd ~A
    absurd A, ~A

next up previous contents
Next: Conclusion-annotated proofs Up: Proof semantics Previous: Conditional proofs   Contents
2004-08-06