Metamath Proof Explorer

Theorem efald

Description: Deduction based on reductio ad absurdum. (Contributed by Mario Carneiro, 9-Feb-2017)

Ref Expression
Hypothesis efald.1 
|- ( ( ph /\ -. ps ) -> F. )
Assertion efald 
|- ( ph -> ps )

Proof

Step Hyp Ref Expression
1 efald.1 
 |-  ( ( ph /\ -. ps ) -> F. )
2 1 inegd 
 |-  ( ph -> -. -. ps )
3 2 notnotrd 
 |-  ( ph -> ps )