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Metamath Proof Explorer

Theorem sbied

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbie ) Usage of this theorem is discouraged because it depends on ax-13 . See sbiedw , sbiedvw for variants using disjoint variables, but requiring fewer axioms. (Contributed by NM, 30-Jun-1994) (Revised by Mario Carneiro, 4-Oct-2016) (Proof shortened by Wolf Lammen, 24-Jun-2018) (New usage is discouraged.)

Ref Expression
Hypotheses sbied.1 
|- F/ x ph
sbied.2 
|- ( ph -> F/ x ch )
sbied.3 
|- ( ph -> ( x = y -> ( ps <-> ch ) ) )
Assertion sbied 
|- ( ph -> ( [ y / x ] ps <-> ch ) )

Proof

Step Hyp Ref Expression
1 sbied.1 
 |-  F/ x ph
2 sbied.2 
 |-  ( ph -> F/ x ch )
3 sbied.3 
 |-  ( ph -> ( x = y -> ( ps <-> ch ) ) )
4 1 sbrim 
 |-  ( [ y / x ] ( ph -> ps ) <-> ( ph -> [ y / x ] ps ) )
5 1 2 nfim1 
 |-  F/ x ( ph -> ch )
6 3 com12 
 |-  ( x = y -> ( ph -> ( ps <-> ch ) ) )
7 6 pm5.74d 
 |-  ( x = y -> ( ( ph -> ps ) <-> ( ph -> ch ) ) )
8 5 7 sbie 
 |-  ( [ y / x ] ( ph -> ps ) <-> ( ph -> ch ) )
9 4 8 bitr3i 
 |-  ( ( ph -> [ y / x ] ps ) <-> ( ph -> ch ) )
10 9 pm5.74ri 
 |-  ( ph -> ( [ y / x ] ps <-> ch ) )