Theorem condandc 882

Description: Proof by contradiction. This only holds for decidable propositions, as it is part of the family of theorems which assume -. ps, derive a contradiction, and therefore conclude ps. By contrast, assuming ps, deriving a contradiction, and therefore concluding -. ps, as in pm2.65 660, is valid for all propositions. (Contributed by Jim Kingdon, 13-May-2018.)

Hypotheses

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

Assertion

Ref	Expression
condandc	|- (DECID ps -> ( ph -> ps ) )

Proof of Theorem condandc

Step	Hyp	Ref	Expression
1		condandc.1	. . 3 |- ( ( ph /\ -. ps ) -> ch )
2		condandc.2	. . 3 |- ( ( ph /\ -. ps ) -> -. ch )
3	1, 2	pm2.65da 662	. 2 |- ( ph -> -. -. ps )
4		notnotrdc 844	. 2 |- (DECID ps -> ( -. -. ps -> ps ) )
5	3, 4	syl5 32	1 |- (DECID ps -> ( ph -> ps ) )

Syntax hints:   -. wn 3   -> wi 4   /\ wa 104  DECID wdc 835

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 710

This theorem depends on definitions:  df-bi 117  df-dc 836

This theorem is referenced by:  ifnetruedc  3603  perfectlem2  15320