Theorem impbidd 126

Description: Deduce an equivalence from two implications. (Contributed by Rodolfo Medina, 12-Oct-2010.)

Hypotheses

Ref	Expression
impbidd.1	|- ( ph -> ( ps -> ( ch -> th ) ) )
impbidd.2	|- ( ph -> ( ps -> ( th -> ch ) ) )

Assertion

Ref	Expression
impbidd	|- ( ph -> ( ps -> ( ch <-> th ) ) )

Proof of Theorem impbidd

Step	Hyp	Ref	Expression
1		impbidd.1	. 2 |- ( ph -> ( ps -> ( ch -> th ) ) )
2		impbidd.2	. 2 |- ( ph -> ( ps -> ( th -> ch ) ) )
3		bi3 118	. 2 |- ( ( ch -> th ) -> ( ( th -> ch ) -> ( ch <-> th ) ) )
4	1, 2, 3	syl6c 66	1 |- ( ph -> ( ps -> ( ch <-> th ) ) )

Syntax hints:   -> wi 4   <-> wb 104

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

This theorem depends on definitions:  df-bi 116

This theorem is referenced by:  impbid21d  127  pm5.74  178  con1biimdc  868  pclem6  1369