Theorem condan 478

Description: Proof by contradiction.

Hypotheses

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

Assertion

Ref	Expression
condan	|- (ph -> ps)

Proof of Theorem condan

Step	Hyp	Ref	Expression
1		condan.1	. . . 4 |- ((ph /\ -. ps) -> ch)
2	1	ex 373	. . 3 |- (ph -> (-. ps -> ch))
3		condan.2	. . . 4 |- ((ph /\ -. ps) -> -. ch)
4	3	ex 373	. . 3 |- (ph -> (-. ps -> -. ch))
5	2, 4	pm2.65d 136	. 2 |- (ph -> -. -. ps)
6		nega 84	. 2 |- (-. -. ps -> ps)
7	5, 6	syl 10	1 |- (ph -> ps)

Syntax hints:  -. wn 2   -> wi 3   /\ wa 223

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 147  df-an 225

Copyright terms: Public domain