Metamath Proof Explorer


Theorem sbequ8

Description: Elimination of equality from antecedent after substitution. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 5-Aug-1993) Reduce dependencies on axioms. (Revised by Wolf Lammen, 28-Jul-2018) Revise df-sb . (Revised by Wolf Lammen, 28-Jul-2023) (New usage is discouraged.)

Ref Expression
Assertion sbequ8
|- ( [ y / x ] ph <-> [ y / x ] ( x = y -> ph ) )

Proof

Step Hyp Ref Expression
1 equsb1
 |-  [ y / x ] x = y
2 1 a1bi
 |-  ( [ y / x ] ph <-> ( [ y / x ] x = y -> [ y / x ] ph ) )
3 sbim
 |-  ( [ y / x ] ( x = y -> ph ) <-> ( [ y / x ] x = y -> [ y / x ] ph ) )
4 2 3 bitr4i
 |-  ( [ y / x ] ph <-> [ y / x ] ( x = y -> ph ) )