Theorem pm2.61dda 768

Description: Elimination of two antecedents. (Contributed by NM, 9-Jul-2013.)

Hypotheses

Ref	Expression
pm2.61dda.1	|- ((ph /\ -. ps) -> th)
pm2.61dda.2	|- ((ph /\ -. ch) -> th)
pm2.61dda.3	|- ((ph /\ (ps /\ ch)) -> th)

Assertion

Ref	Expression
pm2.61dda	|- (ph -> th)

Proof of Theorem pm2.61dda

Step	Hyp	Ref	Expression
1		pm2.61dda.3	. . . 4 |- ((ph /\ (ps /\ ch)) -> th)
2	1	anassrs 629	. . 3 |- (((ph /\ ps) /\ ch) -> th)
3		pm2.61dda.2	. . . 4 |- ((ph /\ -. ch) -> th)
4	3	adantlr 695	. . 3 |- (((ph /\ ps) /\ -. ch) -> th)
5	2, 4	pm2.61dan 766	. 2 |- ((ph /\ ps) -> th)
6		pm2.61dda.1	. 2 |- ((ph /\ -. ps) -> th)
7	5, 6	pm2.61dan 766	1 |- (ph -> th)

This proof depends on syntax axioms:  -. wn 3   -> wi 4   /\ 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: (None)