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 ) )