Metamath Proof Explorer


Theorem elab2gw

Description: Membership in a class abstraction, using two substitution hypotheses to avoid a disjoint variable condition on x and A , which is not usually significant since B is usually a constant. (Contributed by SN, 16-May-2024)

Ref Expression
Hypotheses elabgw.1 ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
elabgw.2 ⊢ ( 𝑦 = 𝐴 → ( 𝜓 ↔ 𝜒 ) )
elab2gw.3 ⊢ 𝐵 = { 𝑥 ∣ 𝜑 }
Assertion elab2gw ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ 𝐵 ↔ 𝜒 ) )

Proof

Step Hyp Ref Expression
1 elabgw.1 ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
2 elabgw.2 ⊢ ( 𝑦 = 𝐴 → ( 𝜓 ↔ 𝜒 ) )
3 elab2gw.3 ⊢ 𝐵 = { 𝑥 ∣ 𝜑 }
4 3 eleq2i ⊢ ( 𝐴 ∈ 𝐵 ↔ 𝐴 ∈ { 𝑥 ∣ 𝜑 } )
5 1 2 elabgw ⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜒 ) )
6 4 5 bitrid ⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ 𝐵 ↔ 𝜒 ) )