Metamath Proof Explorer


Theorem sbc2iedv

Description: Conversion of implicit substitution to explicit class substitution. (Contributed by NM, 16-Dec-2008) (Proof shortened by Mario Carneiro, 18-Oct-2016)

Ref Expression
Hypotheses sbc2iedv.1
|- A e. _V
sbc2iedv.2
|- B e. _V
sbc2iedv.3
|- ( ph -> ( ( x = A /\ y = B ) -> ( ps <-> ch ) ) )
Assertion sbc2iedv
|- ( ph -> ( [. A / x ]. [. B / y ]. ps <-> ch ) )

Proof

Step Hyp Ref Expression
1 sbc2iedv.1
 |-  A e. _V
2 sbc2iedv.2
 |-  B e. _V
3 sbc2iedv.3
 |-  ( ph -> ( ( x = A /\ y = B ) -> ( ps <-> ch ) ) )
4 1 a1i
 |-  ( ph -> A e. _V )
5 2 a1i
 |-  ( ( ph /\ x = A ) -> B e. _V )
6 3 impl
 |-  ( ( ( ph /\ x = A ) /\ y = B ) -> ( ps <-> ch ) )
7 5 6 sbcied
 |-  ( ( ph /\ x = A ) -> ( [. B / y ]. ps <-> ch ) )
8 4 7 sbcied
 |-  ( ph -> ( [. A / x ]. [. B / y ]. ps <-> ch ) )