Metamath Proof Explorer

Theorem sbcor

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

Ref Expression
Assertion sbcor 
|- ( [. 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 ]. ph -> A e. _V )
3 sbcex 
 |-  ( [. A / x ]. ps -> A e. _V )
4 2 3 jaoi 
 |-  ( ( [. A / x ]. ph \/ [. A / x ]. ps ) -> A e. _V )
5 dfsbcq2 
 |-  ( y = A -> ( [ y / x ] ( ph \/ ps ) <-> [. A / x ]. ( ph \/ ps ) ) )
6 dfsbcq2 
 |-  ( y = A -> ( [ y / x ] ph <-> [. A / x ]. ph ) )
7 dfsbcq2 
 |-  ( y = A -> ( [ y / x ] ps <-> [. A / x ]. ps ) )
8 6 7 orbi12d 
 |-  ( y = A -> ( ( [ y / x ] ph \/ [ y / x ] ps ) <-> ( [. A / x ]. ph \/ [. A / x ]. ps ) ) )
9 sbor 
 |-  ( [ y / x ] ( ph \/ ps ) <-> ( [ y / x ] ph \/ [ y / x ] ps ) )
10 5 8 9 vtoclbg 
 |-  ( A e. _V -> ( [. A / x ]. ( ph \/ ps ) <-> ( [. A / x ]. ph \/ [. A / x ]. ps ) ) )
11 1 4 10 pm5.21nii 
 |-  ( [. A / x ]. ( ph \/ ps ) <-> ( [. A / x ]. ph \/ [. A / x ]. ps ) )