Theorem 3impexp 1425

Description: impexp 261 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 1214	. 2 |- ( ( ( ph /\ ps /\ ch ) -> th ) -> ( ph -> ( ps -> ( ch -> th ) ) ) )
3		id 19	. . 3 |- ( ( ph -> ( ps -> ( ch -> th ) ) ) -> ( ph -> ( ps -> ( ch -> th ) ) ) )
4	3	3impd 1211	. 2 |- ( ( ph -> ( ps -> ( ch -> th ) ) ) -> ( ( ph /\ ps /\ ch ) -> th ) )
5	2, 4	impbii 125	1 |- ( ( ( ph /\ ps /\ ch ) -> th ) <-> ( ph -> ( ps -> ( ch -> th ) ) ) )

Syntax hints:   -> wi 4   <-> wb 104   /\ w3a 968

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116  df-3an 970

This theorem is referenced by:  3impexpbicom  1426