Metamath Proof Explorer


Theorem csbie2g

Description: Conversion of implicit substitution to explicit class substitution. This version of csbie avoids a disjointness condition on x , A and x , D by substituting twice. (Contributed by Mario Carneiro, 11-Nov-2016)

Ref Expression
Hypotheses csbie2g.1 ⊢ ( 𝑥 = 𝑦 → 𝐵 = 𝐶 )
csbie2g.2 ⊢ ( 𝑦 = 𝐴 → 𝐶 = 𝐷 )
Assertion csbie2g ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = 𝐷 )

Proof

Step Hyp Ref Expression
1 csbie2g.1 ⊢ ( 𝑥 = 𝑦 → 𝐵 = 𝐶 )
2 csbie2g.2 ⊢ ( 𝑦 = 𝐴 → 𝐶 = 𝐷 )
3 df-csb ⊢ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = { 𝑧 ∣ [ 𝐴 / 𝑥 ] 𝑧 ∈ 𝐵 }
4 1 eleq2d ⊢ ( 𝑥 = 𝑦 → ( 𝑧 ∈ 𝐵 ↔ 𝑧 ∈ 𝐶 ) )
5 2 eleq2d ⊢ ( 𝑦 = 𝐴 → ( 𝑧 ∈ 𝐶 ↔ 𝑧 ∈ 𝐷 ) )
6 4 5 sbcie2g ⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝑧 ∈ 𝐵 ↔ 𝑧 ∈ 𝐷 ) )
7 6 eqabcdv ⊢ ( 𝐴 ∈ 𝑉 → { 𝑧 ∣ [ 𝐴 / 𝑥 ] 𝑧 ∈ 𝐵 } = 𝐷 )
8 3 7 eqtrid ⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = 𝐷 )