Metamath Proof Explorer


Theorem sbcani

Description: Distribution of class substitution over conjunction, in inference form. (Contributed by Giovanni Mascellani, 27-May-2019)

Ref Expression
Hypotheses sbcani.1
|- ( [. A / x ]. ph <-> ch )
sbcani.2
|- ( [. A / x ]. ps <-> et )
Assertion sbcani
|- ( [. A / x ]. ( ph /\ ps ) <-> ( ch /\ et ) )

Proof

Step Hyp Ref Expression
1 sbcani.1
 |-  ( [. A / x ]. ph <-> ch )
2 sbcani.2
 |-  ( [. A / x ]. ps <-> et )
3 sbcan
 |-  ( [. A / x ]. ( ph /\ ps ) <-> ( [. A / x ]. ph /\ [. A / x ]. ps ) )
4 1 2 anbi12i
 |-  ( ( [. A / x ]. ph /\ [. A / x ]. ps ) <-> ( ch /\ et ) )
5 3 4 bitri
 |-  ( [. A / x ]. ( ph /\ ps ) <-> ( ch /\ et ) )