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Metamath Proof Explorer

Theorem sbciegf

Description: Conversion of implicit substitution to explicit class substitution. (Contributed by NM, 14-Dec-2005) (Revised by Mario Carneiro, 13-Oct-2016)

	Ref	Expression
Hypotheses	sbciegf.1	⊢ Ⅎ 𝑥 𝜓
	sbciegf.2	⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
Assertion	sbciegf	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		sbciegf.1	⊢ Ⅎ 𝑥 𝜓
2		sbciegf.2	⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
3	2	ax-gen	⊢ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
4		sbciegft	⊢ ( ( 𝐴 ∈ 𝑉 ∧ Ⅎ 𝑥 𝜓 ∧ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) ) → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝜓 ) )
5	1 3 4	mp3an23	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝜓 ) )