Metamath Proof Explorer


Theorem sbiedv

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbie ). Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker sbiedvw when possible. (Contributed by NM, 7-Jan-2017) (New usage is discouraged.)

Ref Expression
Hypothesis sbiedv.1
|- ( ( ph /\ x = y ) -> ( ps <-> ch ) )
Assertion sbiedv
|- ( ph -> ( [ y / x ] ps <-> ch ) )

Proof

Step Hyp Ref Expression
1 sbiedv.1
 |-  ( ( ph /\ x = y ) -> ( ps <-> ch ) )
2 nfv
 |-  F/ x ph
3 nfvd
 |-  ( ph -> F/ x ch )
4 1 ex
 |-  ( ph -> ( x = y -> ( ps <-> ch ) ) )
5 2 3 4 sbied
 |-  ( ph -> ( [ y / x ] ps <-> ch ) )