Theorem mpjaod 370

Description: Eliminate a disjunction in a deduction. (Contributed by Mario Carneiro, 29-May-2016.)

Hypotheses

Ref	Expression
jaod.1	|- (ph -> (ps -> ch))
jaod.2	|- (ph -> (th -> ch))
jaod.3	|- (ph -> (ps \/ th))

Assertion

Ref	Expression
mpjaod	|- (ph -> ch)

Proof of Theorem mpjaod

Step	Hyp	Ref	Expression
1		jaod.3	. 2 |- (ph -> (ps \/ th))
2		jaod.1	. . 3 |- (ph -> (ps -> ch))
3		jaod.2	. . 3 |- (ph -> (th -> ch))
4	2, 3	jaod 369	. 2 |- (ph -> ((ps \/ th) -> ch))
5	1, 4	mpd 14	1 |- (ph -> ch)

Syntax hints:   -> wi 4   \/ wo 357

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 177  df-or 359

This theorem is referenced by:  ecase2d  906  fnfreclem3  6319