Self-Distributive Law for Conditional/Reverse Implication/Formulation 2

Theorem

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

Proof

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\paren {p \implies q} \implies \paren {p \implies r}$	Assumption	(None)
2		1	$p \implies \paren {q \implies r}$	Sequent Introduction	1	Self-Distributive Law for Conditional: Formulation 1
3			$\paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }$	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): $\text{I}$: 'NOT' and 'IF': $\S 5$: Theorem $\text{T7}$