Metamath Proof Explorer

Theorem renegd

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

Ref Expression
Hypothesis recld.1 
|- ( ph -> A e. CC )
Assertion renegd 
|- ( ph -> ( Re ` -u A ) = -u ( Re ` A ) )

Proof

Step Hyp Ref Expression
1 recld.1 
 |-  ( ph -> A e. CC )
2 reneg 
 |-  ( A e. CC -> ( Re ` -u A ) = -u ( Re ` A ) )
3 1 2 syl 
 |-  ( ph -> ( Re ` -u A ) = -u ( Re ` A ) )