Theorem 3impexp 1326

Description: impexp 250 with a 3-conjunct antecedent. (Contributed by Alan Sare, 31-Dec-2011.)

Assertion

Ref	Expression
3impexp	|- ( ( ( ph /\ ps /\ ch ) -> th ) <-> ( ph -> ( ps -> ( ch -> th ) ) ) )

Proof of Theorem 3impexp

Step	Hyp	Ref	Expression
1		id 19	. . 3 |- ( ( ( ph /\ ps /\ ch ) -> th ) -> ( ( ph /\ ps /\ ch ) -> th ) )
2	1	3expd 1121	. 2 |- ( ( ( ph /\ ps /\ ch ) -> th ) -> ( ph -> ( ps -> ( ch -> th ) ) ) )
3		id 19	. . 3 |- ( ( ph -> ( ps -> ( ch -> th ) ) ) -> ( ph -> ( ps -> ( ch -> th ) ) ) )
4	3	3impd 1118	. 2 |- ( ( ph -> ( ps -> ( ch -> th ) ) ) -> ( ( ph /\ ps /\ ch ) -> th ) )
5	2, 4	impbii 117	1 |- ( ( ( ph /\ ps /\ ch ) -> th ) <-> ( ph -> ( ps -> ( ch -> th ) ) ) )

Syntax hints:   -> wi 4   <-> wb 98   /\ w3a 885

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101

This theorem depends on definitions:  df-bi 110  df-3an 887

This theorem is referenced by:  3impexpbicom  1327