Theorem impbida 805

Description: Deduce an equivalence from two implications. (Contributed by NM, 17-Feb-2007.)

Hypotheses

Ref	Expression
impbida.1	|- ((ph /\ ps) -> ch)
impbida.2	|- ((ph /\ ch) -> ps)

Assertion

Ref	Expression
impbida	|- (ph -> (ps <-> ch))

Proof of Theorem impbida

Step	Hyp	Ref	Expression
1		impbida.1	. . 3 |- ((ph /\ ps) -> ch)
2	1	ex 423	. 2 |- (ph -> (ps -> ch))
3		impbida.2	. . 3 |- ((ph /\ ch) -> ps)
4	3	ex 423	. 2 |- (ph -> (ch -> ps))
5	2, 4	impbid 183	1 |- (ph -> (ps <-> ch))

This proof depends on syntax axioms:   -> wi 4   <-> wb 176   /\ wa 358

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177  df-an 360

This theorem is used by:  eqrdav  2352  funfvbrb  5402  f1o2d  5728  ersymb  5954  erth  5969