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Metamath Proof Explorer

Theorem sbcieg

Description: Conversion of implicit substitution to explicit class substitution. (Contributed by NM, 10-Nov-2005) Avoid ax-10 , ax-12 . (Revised by GG, 12-Oct-2024)

		Ref	Expression
	Hypothesis	sbcieg.1	⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
	Assertion	sbcieg	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		sbcieg.1	⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
2		df-sbc	⊢ ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝐴 ∈ { 𝑥 ∣ 𝜑 } )
3	1	elabg	⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )
4	2 3	bitrid	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝜓 ) )