Theorem pm2.61ddan

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

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

Proof

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