Constructive Dilemma
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Theorem
Formulation 1
\(\ds p \implies q\) | \(\) | \(\ds \) | ||||||||||||
\(\ds r \implies s\) | \(\) | \(\ds \) | ||||||||||||
\(\ds \vdash \ \ \) | \(\ds p \lor r \implies q \lor s\) | \(\) | \(\ds \) |
Formulation 2
\(\ds \paren {p \implies q} \land \paren {r \implies s}\) | \(\) | \(\ds \) | ||||||||||||
\(\ds p \lor r\) | \(\) | \(\ds \) | ||||||||||||
\(\ds \vdash \ \ \) | \(\ds q \lor s\) | \(\) | \(\ds \) |
Formulation 3
- $\vdash \paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}$
Its abbreviation in a tableau proof is $\textrm{CD}$.