Metamath Proof Explorer


Theorem elabgw

Description: Membership in a class abstraction, using implicit substitution and an intermediate setvar y to avoid ax-10 , ax-11 , ax-12 . It also avoids a disjoint variable condition on x and A . This is to elabg what sbievw2 is to sbievw . (Contributed by SN, 20-Apr-2024)

Ref Expression
Hypotheses elabgw.1 ( 𝑥 = 𝑦 → ( 𝜑𝜓 ) )
elabgw.2 ( 𝑦 = 𝐴 → ( 𝜓𝜒 ) )
Assertion elabgw ( 𝐴𝑉 → ( 𝐴 ∈ { 𝑥𝜑 } ↔ 𝜒 ) )

Proof

Step Hyp Ref Expression
1 elabgw.1 ( 𝑥 = 𝑦 → ( 𝜑𝜓 ) )
2 elabgw.2 ( 𝑦 = 𝐴 → ( 𝜓𝜒 ) )
3 eleq1 ( 𝑦 = 𝐴 → ( 𝑦 ∈ { 𝑥𝜑 } ↔ 𝐴 ∈ { 𝑥𝜑 } ) )
4 df-clab ( 𝑦 ∈ { 𝑥𝜑 } ↔ [ 𝑦 / 𝑥 ] 𝜑 )
5 1 sbievw ( [ 𝑦 / 𝑥 ] 𝜑𝜓 )
6 4 5 bitri ( 𝑦 ∈ { 𝑥𝜑 } ↔ 𝜓 )
7 3 2 6 vtoclbg ( 𝐴𝑉 → ( 𝐴 ∈ { 𝑥𝜑 } ↔ 𝜒 ) )