Metamath Proof Explorer

Theorem absnegd

Description: Absolute value of negative. (Contributed by Mario Carneiro, 29-May-2016)

Ref Expression
Hypothesis abscld.1 
|- ( ph -> A e. CC )
Assertion absnegd 
|- ( ph -> ( abs ` -u A ) = ( abs ` A ) )

Proof

Step Hyp Ref Expression
1 abscld.1 
 |-  ( ph -> A e. CC )
2 absneg 
 |-  ( A e. CC -> ( abs ` -u A ) = ( abs ` A ) )
3 1 2 syl 
 |-  ( ph -> ( abs ` -u A ) = ( abs ` A ) )