Metamath Proof Explorer


Theorem condan

Description: Proof by contradiction. (Contributed by NM, 9-Feb-2006) (Proof shortened by Wolf Lammen, 19-Jun-2014)

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

Proof

Step Hyp Ref Expression
1 condan.1
 |-  ( ( ph /\ -. ps ) -> ch )
2 condan.2
 |-  ( ( ph /\ -. ps ) -> -. ch )
3 1 2 pm2.65da
 |-  ( ph -> -. -. ps )
4 3 notnotrd
 |-  ( ph -> ps )