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 $\text{T32}$