Theorem mpjao3dan 1201

Description: Eliminate a 3-way disjunction in a deduction. (Contributed by Thierry Arnoux, 13-Apr-2018.)

Hypotheses

Ref	Expression
mpjao3dan.1	|- ((ph /\ ps) -> ch)
mpjao3dan.2	|- ((ph /\ th) -> ch)
mpjao3dan.3	|- ((ph /\ ta) -> ch)
mpjao3dan.4	|- (ph -> (ps \/ th \/ ta))

Assertion

Ref	Expression
mpjao3dan	|- (ph -> ch)

Proof of Theorem mpjao3dan

Step	Hyp	Ref	Expression
1		mpjao3dan.1	. . 3 |- ((ph /\ ps) -> ch)
2		mpjao3dan.2	. . 3 |- ((ph /\ th) -> ch)
3	1, 2	jaodan 709	. 2 |- ((ph /\ (ps \/ th)) -> ch)
4		mpjao3dan.3	. 2 |- ((ph /\ ta) -> ch)
5		mpjao3dan.4	. . 3 |- (ph -> (ps \/ th \/ ta))
6		df-3or 885	. . 3 |- ((ps \/ th \/ ta) <-> ((ps \/ th) \/ ta))
7	5, 6	sylib 127	. 2 |- (ph -> ((ps \/ th) \/ ta))
8	3, 4, 7	mpjaodan 710	1 |- (ph -> ch)

Syntax hints:   -> wi 4   /\ wa 97   \/ wo 628   \/ w3o 883

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629

This theorem depends on definitions:  df-bi 110  df-3or 885

This theorem is referenced by:  nntri3  6014  caucvgprlemnkj  6637  caucvgprlemnbj  6638