Metamath Proof Explorer


Theorem cdeqnot

Description: Distribute conditional equality over negation. (Contributed by Mario Carneiro, 11-Aug-2016)

Ref Expression
Hypothesis cdeqnot.1
|- CondEq ( x = y -> ( ph <-> ps ) )
Assertion cdeqnot
|- CondEq ( x = y -> ( -. ph <-> -. ps ) )

Proof

Step Hyp Ref Expression
1 cdeqnot.1
 |-  CondEq ( x = y -> ( ph <-> ps ) )
2 1 cdeqri
 |-  ( x = y -> ( ph <-> ps ) )
3 2 notbid
 |-  ( x = y -> ( -. ph <-> -. ps ) )
4 3 cdeqi
 |-  CondEq ( x = y -> ( -. ph <-> -. ps ) )