Theorem sbc6gOLD

Description: Obsolete version of sbc6g as of 5-Oct-2024. (Contributed by NM, 11-Oct-2004) (Proof shortened by Andrew Salmon, 8-Jun-2011) (Proof modification is discouraged.) (New usage is discouraged.)

	Ref	Expression
Assertion	sbc6gOLD	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ) )

Proof

Step	Hyp	Ref	Expression
1		sbc5	⊢ ( [ 𝐴 / 𝑥 ] 𝜑 ↔ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) )
2		alexeqg	⊢ ( 𝐴 ∈ 𝑉 → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ↔ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ) )
3	1 2	bitr4id	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ) )