Constructive Dilemma/Formulation 2

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Theorem

\(\ds \paren {p \implies q} \land \paren {r \implies s}\) \(\) \(\ds \)
\(\ds p \lor r\) \(\) \(\ds \)
\(\ds \vdash \ \ \) \(\ds q \lor s\) \(\) \(\ds \)


Proof

By the tableau method of natural deduction:

$\paren {p \implies q} \land \paren {r \implies s}, p \lor r \vdash q \lor s$
Line Pool Formula Rule Depends upon Notes
1 1 $\paren {p \implies q} \land \paren {r \implies s}$ Premise (None)
2 2 $p \lor r$ Premise (None)
3 1, 2 $\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s}$ Rule of Conjunction: $\land \II$ 2, 1 Associativity is implicit
4 $\paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}$ Theorem Introduction (None) Constructive Dilemma: Formulation 3
5 1, 2 $q \lor s$ Modus Ponendo Ponens: $\implies \mathcal E$ 4, 3

$\blacksquare$


Sources