Metamath Proof Explorer


Theorem sbcel1g

Description: Move proper substitution in and out of a membership relation. Note that the scope of [. A / x ]. is the wff B e. C , whereas the scope of [_ A / x ]_ is the class B . (Contributed by NM, 10-Nov-2005)

Ref Expression
Assertion sbcel1g
|- ( A e. V -> ( [. A / x ]. B e. C <-> [_ A / x ]_ B e. C ) )

Proof

Step Hyp Ref Expression
1 sbcel12
 |-  ( [. A / x ]. B e. C <-> [_ A / x ]_ B e. [_ A / x ]_ C )
2 csbconstg
 |-  ( A e. V -> [_ A / x ]_ C = C )
3 2 eleq2d
 |-  ( A e. V -> ( [_ A / x ]_ B e. [_ A / x ]_ C <-> [_ A / x ]_ B e. C ) )
4 1 3 bitrid
 |-  ( A e. V -> ( [. A / x ]. B e. C <-> [_ A / x ]_ B e. C ) )