Metamath Proof Explorer


Theorem sbciedf

Description: Conversion of implicit substitution to explicit class substitution, deduction form. (Contributed by NM, 29-Dec-2014)

Ref Expression
Hypotheses sbcied.1
|- ( ph -> A e. V )
sbcied.2
|- ( ( ph /\ x = A ) -> ( ps <-> ch ) )
sbciedf.3
|- F/ x ph
sbciedf.4
|- ( ph -> F/ x ch )
Assertion sbciedf
|- ( ph -> ( [. A / x ]. ps <-> ch ) )

Proof

Step Hyp Ref Expression
1 sbcied.1
 |-  ( ph -> A e. V )
2 sbcied.2
 |-  ( ( ph /\ x = A ) -> ( ps <-> ch ) )
3 sbciedf.3
 |-  F/ x ph
4 sbciedf.4
 |-  ( ph -> F/ x ch )
5 2 ex
 |-  ( ph -> ( x = A -> ( ps <-> ch ) ) )
6 3 5 alrimi
 |-  ( ph -> A. x ( x = A -> ( ps <-> ch ) ) )
7 sbciegft
 |-  ( ( A e. V /\ F/ x ch /\ A. x ( x = A -> ( ps <-> ch ) ) ) -> ( [. A / x ]. ps <-> ch ) )
8 1 4 6 7 syl3anc
 |-  ( ph -> ( [. A / x ]. ps <-> ch ) )