Metamath Proof Explorer


Theorem sbcieg

Description: Conversion of implicit substitution to explicit class substitution. (Contributed by NM, 10-Nov-2005) Avoid ax-10 , ax-12 . (Revised by GG, 12-Oct-2024)

Ref Expression
Hypothesis sbcieg.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
Assertion sbcieg ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝜓 ) )

Proof

Step Hyp Ref Expression
1 sbcieg.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
2 df-sbc ⊢ ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝐴 ∈ { 𝑥 ∣ 𝜑 } )
3 1 elabg ⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )
4 2 3 bitrid ⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝜓 ) )