Metamath Proof Explorer


Theorem sbc6g

Description: An equivalence for class substitution. (Contributed by NM, 11-Oct-2004) (Proof shortened by Andrew Salmon, 8-Jun-2011) (Proof shortened by SN, 5-Oct-2024)

Ref Expression
Assertion sbc6g ( 𝐴𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝐴𝜑 ) ) )

Proof

Step Hyp Ref Expression
1 df-sbc ( [ 𝐴 / 𝑥 ] 𝜑𝐴 ∈ { 𝑥𝜑 } )
2 elab6g ( 𝐴𝑉 → ( 𝐴 ∈ { 𝑥𝜑 } ↔ ∀ 𝑥 ( 𝑥 = 𝐴𝜑 ) ) )
3 1 2 syl5bb ( 𝐴𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝐴𝜑 ) ) )