Metamath Proof Explorer


Theorem sbc2iedf

Description: Conversion of implicit substitution to explicit class substitution. (Contributed by Thierry Arnoux, 4-Jul-2023)

Ref Expression
Hypotheses sbc2iedf.1 ⊢ Ⅎ 𝑥 𝜑
sbc2iedf.2 ⊢ Ⅎ 𝑦 𝜑
sbc2iedf.3 ⊢ Ⅎ 𝑥 𝜒
sbc2iedf.4 ⊢ Ⅎ 𝑦 𝜒
sbc2iedf.5 ⊢ ( 𝜑 → 𝐴 ∈ 𝑉 )
sbc2iedf.6 ⊢ ( 𝜑 → 𝐵 ∈ 𝑊 )
sbc2iedf.7 ⊢ ( ( 𝜑 ∧ ( 𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ) ) → ( 𝜓 ↔ 𝜒 ) )
Assertion sbc2iedf ( 𝜑 → ( [ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝜓 ↔ 𝜒 ) )

Proof

Step Hyp Ref Expression
1 sbc2iedf.1 ⊢ Ⅎ 𝑥 𝜑
2 sbc2iedf.2 ⊢ Ⅎ 𝑦 𝜑
3 sbc2iedf.3 ⊢ Ⅎ 𝑥 𝜒
4 sbc2iedf.4 ⊢ Ⅎ 𝑦 𝜒
5 sbc2iedf.5 ⊢ ( 𝜑 → 𝐴 ∈ 𝑉 )
6 sbc2iedf.6 ⊢ ( 𝜑 → 𝐵 ∈ 𝑊 )
7 sbc2iedf.7 ⊢ ( ( 𝜑 ∧ ( 𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ) ) → ( 𝜓 ↔ 𝜒 ) )
8 6 adantr ⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → 𝐵 ∈ 𝑊 )
9 7 anassrs ⊢ ( ( ( 𝜑 ∧ 𝑥 = 𝐴 ) ∧ 𝑦 = 𝐵 ) → ( 𝜓 ↔ 𝜒 ) )
10 nfv ⊢ Ⅎ 𝑦 𝑥 = 𝐴
11 2 10 nfan ⊢ Ⅎ 𝑦 ( 𝜑 ∧ 𝑥 = 𝐴 )
12 4 a1i ⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → Ⅎ 𝑦 𝜒 )
13 8 9 11 12 sbciedf ⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( [ 𝐵 / 𝑦 ] 𝜓 ↔ 𝜒 ) )
14 3 a1i ⊢ ( 𝜑 → Ⅎ 𝑥 𝜒 )
15 5 13 1 14 sbciedf ⊢ ( 𝜑 → ( [ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝜓 ↔ 𝜒 ) )