Exportation and Self-Conditional

Prove Shunting: $p \land q \Rightarrow r \equiv ( p \Rightarrow q ) \Rightarrow ( p \Rightarrow r)$

Proof: $( p \Rightarrow q ) \Rightarrow ( p \Rightarrow r)$

=

$\neg(\neg p \lor q) \lor (\neg p \lor r)$

=

$(p \land \neg q) \lor (\neg p \land r)$

=

$(p \lor \neg p \lor r) \land (\neg q \lor \neg p \lor r)$

=

$(True \lor r) \land (\neg q \lor \neg p \lor r)$

=

$(\neg q \lor \neg p \lor r)$

=

$p \land q \Rightarrow r$