Theorem ecase3 559

Description: Inference for elimination by cases.

Hypotheses

Ref	Expression
ecase3.1	|- (ph -> ch)
ecase3.2	|- (ps -> ch)
ecase3.3	|- (-. (ph \/ ps) -> ch)

Assertion

Ref	Expression
ecase3	|- ch

Proof of Theorem ecase3

Step	Hyp	Ref	Expression
1		ioran 254	. . . 4 |- (-. (ph \/ ps) <-> (-. ph /\ -. ps))
2		ecase3.3	. . . 4 |- (-. (ph \/ ps) -> ch)
3	1, 2	sylbir 176	. . 3 |- ((-. ph /\ -. ps) -> ch)
4	3	exp 291	. 2 |- (-. ph -> (-. ps -> ch))
5		ecase3.1	. 2 |- (ph -> ch)
6		ecase3.2	. 2 |- (ps -> ch)
7	4, 5, 6	pm2.61ii 113	1 |- ch

Syntax hints:  -. wn 1   -> wi 2   \/ wo 195   /\ wa 196

This theorem is referenced by:  ecase3d 560  eueq3 1430

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198

metamath.org