Theorem sbequ8

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.)

Conclusion
\( \vdash [ y' / x ] \, \varphi \leftrightarrow [ y' / x ] \, (x = y' \rightarrow \varphi) \)
Proof
equsb1
l1
a1bi _
\( \vdash [ y' / x ] \, \varphi \leftrightarrow [ y' / x ] \, x = y' \rightarrow [ y' / x ] \, \varphi \)
sbim
\( \vdash [ y' / x ] \, (x = y' \rightarrow \varphi) \leftrightarrow [ y' / x ] \, x = y' \rightarrow [ y' / x ] \, \varphi \)
bitr4i l1 _