Metamath Proof Explorer

Theorem sbcan

Description: Distribution of class substitution over conjunction. (Contributed by NM, 31-Dec-2016) (Revised by NM, 17-Aug-2018)

Ref Expression
Assertion sbcan 
|- ( [. A / x ]. ( ph /\ ps ) <-> ( [. A / x ]. ph /\ [. A / x ]. ps ) )

Proof

Step Hyp Ref Expression
1 sbcex 
 |-  ( [. A / x ]. ( ph /\ ps ) -> A e. _V )
2 sbcex 
 |-  ( [. A / x ]. ps -> A e. _V )
3 2 adantl 
 |-  ( ( [. A / x ]. ph /\ [. A / x ]. ps ) -> A e. _V )
4 dfsbcq2 
 |-  ( y = A -> ( [ y / x ] ( ph /\ ps ) <-> [. A / x ]. ( ph /\ ps ) ) )
5 dfsbcq2 
 |-  ( y = A -> ( [ y / x ] ph <-> [. A / x ]. ph ) )
6 dfsbcq2 
 |-  ( y = A -> ( [ y / x ] ps <-> [. A / x ]. ps ) )
7 5 6 anbi12d 
 |-  ( y = A -> ( ( [ y / x ] ph /\ [ y / x ] ps ) <-> ( [. A / x ]. ph /\ [. A / x ]. ps ) ) )
8 sban 
 |-  ( [ y / x ] ( ph /\ ps ) <-> ( [ y / x ] ph /\ [ y / x ] ps ) )
9 4 7 8 vtoclbg 
 |-  ( A e. _V -> ( [. A / x ]. ( ph /\ ps ) <-> ( [. A / x ]. ph /\ [. A / x ]. ps ) ) )
10 1 3 9 pm5.21nii 
 |-  ( [. A / x ]. ( ph /\ ps ) <-> ( [. A / x ]. ph /\ [. A / x ]. ps ) )