Theorem ecase3d

Description: Deduction for elimination by cases. (Contributed by NM, 2-May-1996) (Proof shortened by Andrew Salmon, 7-May-2011)

	Ref	Expression
Hypotheses	ecase3d.1	|- ( ph -> ( ps -> th ) )
	ecase3d.2	|- ( ph -> ( ch -> th ) )
	ecase3d.3	|- ( ph -> ( -. ( ps \/ ch ) -> th ) )
Assertion	ecase3d	|- ( ph -> th )

Proof

Step	Hyp	Ref	Expression
1		ecase3d.1	|- ( ph -> ( ps -> th ) )
2		ecase3d.2	|- ( ph -> ( ch -> th ) )
3		ecase3d.3	|- ( ph -> ( -. ( ps \/ ch ) -> th ) )
4	1 2	jaod	|- ( ph -> ( ( ps \/ ch ) -> th ) )
5	4 3	pm2.61d	|- ( ph -> th )