Rule of Exportation/Forward Implication/Formulation 2

Theorem

$\vdash \left({\left ({p \land q}\right) \implies r}\right) \implies \left({p \implies \left ({q \implies r}\right)}\right)$

Proof 1

By the tableau method of natural deduction:

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

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

$\blacksquare$

Proof 2

We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations.

$\begin{array}{|ccccc|c|ccccc|} \hline ((p & \land & q) & \implies & r) & \implies & (p & \implies & (q & \implies & r)) \\ \hline F & F & F & T & F & T & F & T & F & T & F \\ F & F & F & T & T & T & F & T & F & T & T \\ F & F & T & T & F & T & F & T & T & F & F \\ F & F & T & T & T & T & F & T & T & T & T \\ T & F & F & T & F & T & T & T & F & T & F \\ T & F & F & T & T & T & T & T & F & T & T \\ T & T & T & F & F & T & T & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$

$\blacksquare$