Metamath Proof Explorer


Theorem csbov12g

Description: Move class substitution in and out of an operation. (Contributed by NM, 12-Nov-2005)

Ref Expression
Assertion csbov12g
|- ( A e. V -> [_ A / x ]_ ( B F C ) = ( [_ A / x ]_ B F [_ A / x ]_ C ) )

Proof

Step Hyp Ref Expression
1 csbov123
 |-  [_ A / x ]_ ( B F C ) = ( [_ A / x ]_ B [_ A / x ]_ F [_ A / x ]_ C )
2 csbconstg
 |-  ( A e. V -> [_ A / x ]_ F = F )
3 2 oveqd
 |-  ( A e. V -> ( [_ A / x ]_ B [_ A / x ]_ F [_ A / x ]_ C ) = ( [_ A / x ]_ B F [_ A / x ]_ C ) )
4 1 3 eqtrid
 |-  ( A e. V -> [_ A / x ]_ ( B F C ) = ( [_ A / x ]_ B F [_ A / x ]_ C ) )