Metamath Proof Explorer


Theorem sbcexf

Description: Move existential quantifier in and out of class substitution, with an explicit nonfree variable condition. (Contributed by Giovanni Mascellani, 29-May-2019)

Ref Expression
Hypothesis sbcexf.1
|- F/_ y A
Assertion sbcexf
|- ( [. A / x ]. E. y ph <-> E. y [. A / x ]. ph )

Proof

Step Hyp Ref Expression
1 sbcexf.1
 |-  F/_ y A
2 nfv
 |-  F/ z ph
3 2 sb8ev
 |-  ( E. y ph <-> E. z [ z / y ] ph )
4 3 sbcbii
 |-  ( [. A / x ]. E. y ph <-> [. A / x ]. E. z [ z / y ] ph )
5 sbcex2
 |-  ( [. A / x ]. E. z [ z / y ] ph <-> E. z [. A / x ]. [ z / y ] ph )
6 nfs1v
 |-  F/ y [ z / y ] ph
7 1 6 nfsbcw
 |-  F/ y [. A / x ]. [ z / y ] ph
8 nfv
 |-  F/ z [. A / x ]. ph
9 sbequ12r
 |-  ( z = y -> ( [ z / y ] ph <-> ph ) )
10 9 sbcbidv
 |-  ( z = y -> ( [. A / x ]. [ z / y ] ph <-> [. A / x ]. ph ) )
11 7 8 10 cbvexv1
 |-  ( E. z [. A / x ]. [ z / y ] ph <-> E. y [. A / x ]. ph )
12 4 5 11 3bitri
 |-  ( [. A / x ]. E. y ph <-> E. y [. A / x ]. ph )