Theorem ecase2d 824

Description: Deduction for elimination by cases.

Hypotheses

Ref	Expression
ecase2d.1	|- (ph -> ps)
ecase2d.2	|- (ph -> -. (ps /\ ch))
ecase2d.3	|- (ph -> -. (ps /\ th))
ecase2d.4	|- (ph -> (ta \/ (ch \/ th)))

Assertion

Ref	Expression
ecase2d	|- (ph -> ta)

Proof of Theorem ecase2d

Step	Hyp	Ref	Expression
1		ioran 331	. . 3 |- (-. (ch \/ th) <-> (-. ch /\ -. th))
2		ecase2d.1	. . . 4 |- (ph -> ps)
3		ecase2d.2	. . . . 5 |- (ph -> -. (ps /\ ch))
4		imnan 261	. . . . 5 |- ((ps -> -. ch) <-> -. (ps /\ ch))
5	3, 4	sylibr 217	. . . 4 |- (ph -> (ps -> -. ch))
6	2, 5	mpd 29	. . 3 |- (ph -> -. ch)
7		ecase2d.3	. . . . 5 |- (ph -> -. (ps /\ th))
8		imnan 261	. . . . 5 |- ((ps -> -. th) <-> -. (ps /\ th))
9	7, 8	sylibr 217	. . . 4 |- (ph -> (ps -> -. th))
10	2, 9	mpd 29	. . 3 |- (ph -> -. th)
11	1, 6, 10	sylanbrc 527	. 2 |- (ph -> -. (ch \/ th))
12		ecase2d.4	. . . 4 |- (ph -> (ta \/ (ch \/ th)))
13		orcom 266	. . . 4 |- ((ta \/ (ch \/ th)) <-> ((ch \/ th) \/ ta))
14	12, 13	sylib 215	. . 3 |- (ph -> ((ch \/ th) \/ ta))
15	14	ord 249	. 2 |- (ph -> (-. (ch \/ th) -> ta))
16	11, 15	mpd 29	1 |- (ph -> ta)

Syntax hints:  -. wn 2   -> wi 3   \/ wo 239   /\ wa 240

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7

This theorem depends on definitions:  df-bi 164  df-or 241  df-an 242

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