Metamath Proof Explorer


Theorem csbnegg

Description: Move class substitution in and out of the negative of a number. (Contributed by NM, 1-Mar-2008) (Proof shortened by Andrew Salmon, 22-Oct-2011)

Ref Expression
Assertion csbnegg
|- ( A e. V -> [_ A / x ]_ -u B = -u [_ A / x ]_ B )

Proof

Step Hyp Ref Expression
1 csbov2g
 |-  ( A e. V -> [_ A / x ]_ ( 0 - B ) = ( 0 - [_ A / x ]_ B ) )
2 df-neg
 |-  -u B = ( 0 - B )
3 2 csbeq2i
 |-  [_ A / x ]_ -u B = [_ A / x ]_ ( 0 - B )
4 df-neg
 |-  -u [_ A / x ]_ B = ( 0 - [_ A / x ]_ B )
5 1 3 4 3eqtr4g
 |-  ( A e. V -> [_ A / x ]_ -u B = -u [_ A / x ]_ B )