Theorem ecase2d 558

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		ecase2d.1	. . . . 5 |- (ph -> ps)
2		ecase2d.2	. . . . . 6 |- (ph -> -. (ps /\ ch))
3		imnan 207	. . . . . 6 |- ((ps -> -. ch) <-> -. (ps /\ ch))
4	2, 3	sylibr 175	. . . . 5 |- (ph -> (ps -> -. ch))
5	1, 4	mpd 46	. . . 4 |- (ph -> -. ch)
6		ecase2d.3	. . . . . 6 |- (ph -> -. (ps /\ th))
7		imnan 207	. . . . . 6 |- ((ps -> -. th) <-> -. (ps /\ th))
8	6, 7	sylibr 175	. . . . 5 |- (ph -> (ps -> -. th))
9	1, 8	mpd 46	. . . 4 |- (ph -> -. th)
10	5, 9	jca 236	. . 3 |- (ph -> (-. ch /\ -. th))
11		ioran 254	. . 3 |- (-. (ch \/ th) <-> (-. ch /\ -. th))
12	10, 11	sylibr 175	. 2 |- (ph -> -. (ch \/ th))
13		ecase2d.4	. . . 4 |- (ph -> (ta \/ (ch \/ th)))
14		orcom 209	. . . 4 |- ((ta \/ (ch \/ th)) <-> ((ch \/ th) \/ ta ))
15	13, 14	sylib 173	. . 3 |- (ph -> ((ch \/ th) \/ ta ))
16	15	ord 202	. 2 |- (ph -> (-. (ch \/ th) -> ta ))
17	12, 16	mpd 46	1 |- (ph -> ta )

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

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

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