Metamath Proof Explorer


Theorem sbiedvw

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbievw ). Version of sbied and sbiedv with more disjoint variable conditions, requiring fewer axioms. (Contributed by NM, 30-Jun-1994) (Revised by Gino Giotto, 29-Jan-2024)

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

Proof

Step Hyp Ref Expression
1 sbiedvw.1
 |-  ( ( ph /\ x = y ) -> ( ps <-> ch ) )
2 sbrimvw
 |-  ( [ y / x ] ( ph -> ps ) <-> ( ph -> [ y / x ] ps ) )
3 1 expcom
 |-  ( x = y -> ( ph -> ( ps <-> ch ) ) )
4 3 pm5.74d
 |-  ( x = y -> ( ( ph -> ps ) <-> ( ph -> ch ) ) )
5 4 sbievw
 |-  ( [ y / x ] ( ph -> ps ) <-> ( ph -> ch ) )
6 2 5 bitr3i
 |-  ( ( ph -> [ y / x ] ps ) <-> ( ph -> ch ) )
7 6 pm5.74ri
 |-  ( ph -> ( [ y / x ] ps <-> ch ) )