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Metamath Proof Explorer

Theorem sbc6g

Description: An equivalence for class substitution. (Contributed by NM, 11-Oct-2004) (Proof shortened by Andrew Salmon, 8-Jun-2011) (Proof shortened by SN, 5-Oct-2024)

		Ref	Expression
	Assertion	sbc6g	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ) )

Proof

Step	Hyp	Ref	Expression
1		df-sbc	⊢ ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝐴 ∈ { 𝑥 ∣ 𝜑 } )
2		elab6g	⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ) )
3	1 2	bitrid	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ) )