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Metamath Proof Explorer

Theorem sbie

Description: Conversion of implicit substitution to explicit substitution. For versions requiring disjoint variables, but fewer axioms, see sbiev and sbievw . Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 30-Jun-1994) (Revised by Mario Carneiro, 4-Oct-2016) (Proof shortened by Wolf Lammen, 13-Jul-2019) (New usage is discouraged.)

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

Proof

Step Hyp Ref Expression
1 sbie.1 
 |-  F/ x ps
2 sbie.2 
 |-  ( x = y -> ( ph <-> ps ) )
3 equsb1 
 |-  [ y / x ] x = y
4 2 sbimi 
 |-  ( [ y / x ] x = y -> [ y / x ] ( ph <-> ps ) )
5 3 4 ax-mp 
 |-  [ y / x ] ( ph <-> ps )
6 1 sbf 
 |-  ( [ y / x ] ps <-> ps )
7 6 sblbis 
 |-  ( [ y / x ] ( ph <-> ps ) <-> ( [ y / x ] ph <-> ps ) )
8 5 7 mpbi 
 |-  ( [ y / x ] ph <-> ps )