Metamath Proof Explorer


Theorem elab3gf

Description: Membership in a class abstraction, with a weaker antecedent than elabgf . (Contributed by NM, 6-Sep-2011)

Ref Expression
Hypotheses elab3gf.1 𝑥 𝐴
elab3gf.2 𝑥 𝜓
elab3gf.3 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
Assertion elab3gf ( ( 𝜓𝐴𝐵 ) → ( 𝐴 ∈ { 𝑥𝜑 } ↔ 𝜓 ) )

Proof

Step Hyp Ref Expression
1 elab3gf.1 𝑥 𝐴
2 elab3gf.2 𝑥 𝜓
3 elab3gf.3 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
4 1 2 3 elabgf ( 𝐴 ∈ { 𝑥𝜑 } → ( 𝐴 ∈ { 𝑥𝜑 } ↔ 𝜓 ) )
5 4 ibi ( 𝐴 ∈ { 𝑥𝜑 } → 𝜓 )
6 pm2.21 ( ¬ 𝜓 → ( 𝜓𝐴 ∈ { 𝑥𝜑 } ) )
7 5 6 impbid2 ( ¬ 𝜓 → ( 𝐴 ∈ { 𝑥𝜑 } ↔ 𝜓 ) )
8 1 2 3 elabgf ( 𝐴𝐵 → ( 𝐴 ∈ { 𝑥𝜑 } ↔ 𝜓 ) )
9 7 8 ja ( ( 𝜓𝐴𝐵 ) → ( 𝐴 ∈ { 𝑥𝜑 } ↔ 𝜓 ) )