Metamath Proof Explorer


Theorem sbciegf

Description: Conversion of implicit substitution to explicit class substitution. (Contributed by NM, 14-Dec-2005) (Revised by Mario Carneiro, 13-Oct-2016)

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

Proof

Step Hyp Ref Expression
1 sbciegf.1 ⊢ Ⅎ 𝑥 𝜓
2 sbciegf.2 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
3 2 ax-gen ⊢ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
4 sbciegft ⊢ ( ( 𝐴 ∈ 𝑉 ∧ Ⅎ 𝑥 𝜓 ∧ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) ) → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝜓 ) )
5 1 3 4 mp3an23 ⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝜓 ) )