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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 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] ( 𝑥 = 𝑦 → 𝜑 ) )

Proof

Step	Hyp	Ref	Expression
1		equsb1	⊢ [ 𝑦 / 𝑥 ] 𝑥 = 𝑦
2	1	a1bi	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ( [ 𝑦 / 𝑥 ] 𝑥 = 𝑦 → [ 𝑦 / 𝑥 ] 𝜑 ) )
3		sbim	⊢ ( [ 𝑦 / 𝑥 ] ( 𝑥 = 𝑦 → 𝜑 ) ↔ ( [ 𝑦 / 𝑥 ] 𝑥 = 𝑦 → [ 𝑦 / 𝑥 ] 𝜑 ) )
4	2 3	bitr4i	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] ( 𝑥 = 𝑦 → 𝜑 ) )