Theorem pm5.21nd 866

Description: Eliminate an antecedent implied by each side of a biconditional. (Contributed by NM, 20-Nov-2005.) (Proof shortened by Wolf Lammen, 4-Nov-2013.)

Hypotheses

Ref	Expression
pm5.21nd.1	|- ( ( ph /\ ps ) -> th )
pm5.21nd.2	|- ( ( ph /\ ch ) -> th )
pm5.21nd.3	|- ( th -> ( ps <-> ch ) )

Assertion

Ref	Expression
pm5.21nd	|- ( ph -> ( ps <-> ch ) )

Proof of Theorem pm5.21nd

Step	Hyp	Ref	Expression
1		pm5.21nd.1	. . 3 |- ( ( ph /\ ps ) -> th )
2	1	ex 114	. 2 |- ( ph -> ( ps -> th ) )
3		pm5.21nd.2	. . 3 |- ( ( ph /\ ch ) -> th )
4	3	ex 114	. 2 |- ( ph -> ( ch -> th ) )
5		pm5.21nd.3	. . 3 |- ( th -> ( ps <-> ch ) )
6	5	a1i 9	. 2 |- ( ph -> ( th -> ( ps <-> ch ) ) )
7	2, 4, 6	pm5.21ndd 659	1 |- ( ph -> ( ps <-> ch ) )

Syntax hints:   -> wi 4   /\ wa 103   <-> wb 104

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

This theorem depends on definitions:  df-bi 116

This theorem is referenced by:  ideqg  4618  fvelimab  5395  releldm2  5993  relelec  6372  fzrev3  9650  elfzp12  9662  eltg  11919  eltg2  11920  cncnp2m  12097