Metamath Proof Explorer


Theorem sbc7

Description: An equivalence for class substitution in the spirit of df-clab . Note that x and A don't have to be distinct. (Contributed by NM, 18-Nov-2008) (Revised by Mario Carneiro, 13-Oct-2016)

Ref Expression
Assertion sbc7 ( [ 𝐴 / 𝑥 ] 𝜑 ↔ ∃ 𝑦 ( 𝑦 = 𝐴[ 𝑦 / 𝑥 ] 𝜑 ) )

Proof

Step Hyp Ref Expression
1 sbccow ( [ 𝐴 / 𝑦 ] [ 𝑦 / 𝑥 ] 𝜑[ 𝐴 / 𝑥 ] 𝜑 )
2 sbc5 ( [ 𝐴 / 𝑦 ] [ 𝑦 / 𝑥 ] 𝜑 ↔ ∃ 𝑦 ( 𝑦 = 𝐴[ 𝑦 / 𝑥 ] 𝜑 ) )
3 1 2 bitr3i ( [ 𝐴 / 𝑥 ] 𝜑 ↔ ∃ 𝑦 ( 𝑦 = 𝐴[ 𝑦 / 𝑥 ] 𝜑 ) )