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Metamath Proof Explorer

Theorem sbcie2g

Description: Conversion of implicit substitution to explicit class substitution. This version of sbcie avoids a disjointness condition on x , A by substituting twice. (Contributed by Mario Carneiro, 15-Oct-2016)

	Ref	Expression
Hypotheses	sbcie2g.1	⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
	sbcie2g.2	⊢ ( 𝑦 = 𝐴 → ( 𝜓 ↔ 𝜒 ) )
Assertion	sbcie2g	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		sbcie2g.1	⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
2		sbcie2g.2	⊢ ( 𝑦 = 𝐴 → ( 𝜓 ↔ 𝜒 ) )
3		dfsbcq	⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝐴 / 𝑥 ] 𝜑 ) )
4		sbsbc	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] 𝜑 )
5	1	sbievw	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ 𝜓 )
6	4 5	bitr3i	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ 𝜓 )
7	3 2 6	vtoclbg	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝜒 ) )