Praeclarum Theorema/Formulation 2

Theorem

$\vdash \paren {\paren {p \implies q} \land \paren {r \implies s} } \implies \paren {\paren {p \land r} \implies \paren {q \land s} }$


Proof

By the tableau method of natural deduction:

$\vdash \paren {\paren {p \implies q} \land \paren {r \implies s} } \implies \paren {\paren {p \land r} \implies \paren {q \land s} } $
Line Pool Formula Rule Depends upon Notes
1 1 $\paren {\paren {p \implies q} \land \paren {r \implies s} }$ Assumption (None)
2 1 $\paren {p \land r} \implies \paren {q \land s}$ Sequent Introduction 1 Praeclarum Theorema: Formulation 1
3 1 $\paren {\paren {p \implies q} \land \paren {r \implies s} } \implies \paren {\paren {p \land r} \implies \paren {q \land s} }$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged

$\blacksquare$


Sources

  • 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$: Theorem $\text{T32}$
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