Metamath Proof Explorer


Theorem sbciedf

Description: Conversion of implicit substitution to explicit class substitution, deduction form. (Contributed by NM, 29-Dec-2014)

Ref Expression
Hypotheses sbcied.1 ⊢ ( 𝜑 → 𝐴 ∈ 𝑉 )
sbcied.2 ⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( 𝜓 ↔ 𝜒 ) )
sbciedf.3 ⊢ Ⅎ 𝑥 𝜑
sbciedf.4 ⊢ ( 𝜑 → Ⅎ 𝑥 𝜒 )
Assertion sbciedf ( 𝜑 → ( [ 𝐴 / 𝑥 ] 𝜓 ↔ 𝜒 ) )

Proof

Step Hyp Ref Expression
1 sbcied.1 ⊢ ( 𝜑 → 𝐴 ∈ 𝑉 )
2 sbcied.2 ⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( 𝜓 ↔ 𝜒 ) )
3 sbciedf.3 ⊢ Ⅎ 𝑥 𝜑
4 sbciedf.4 ⊢ ( 𝜑 → Ⅎ 𝑥 𝜒 )
5 2 ex ⊢ ( 𝜑 → ( 𝑥 = 𝐴 → ( 𝜓 ↔ 𝜒 ) ) )
6 3 5 alrimi ⊢ ( 𝜑 → ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜓 ↔ 𝜒 ) ) )
7 sbciegft ⊢ ( ( 𝐴 ∈ 𝑉 ∧ Ⅎ 𝑥 𝜒 ∧ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜓 ↔ 𝜒 ) ) ) → ( [ 𝐴 / 𝑥 ] 𝜓 ↔ 𝜒 ) )
8 1 4 6 7 syl3anc ⊢ ( 𝜑 → ( [ 𝐴 / 𝑥 ] 𝜓 ↔ 𝜒 ) )