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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 ) )