Theorem pm5.21nd 868

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 423	. 2 |- (ph -> (ps -> th))
3		pm5.21nd.2	. . 3 |- ((ph /\ ch) -> th)
4	3	ex 423	. 2 |- (ph -> (ch -> th))
5		pm5.21nd.3	. . 3 |- (th -> (ps <-> ch))
6	5	a1i 10	. 2 |- (ph -> (th -> (ps <-> ch)))
7	2, 4, 6	pm5.21ndd 343	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:  ideqg  4869  ideqg2  4870  fvelimab  5371