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Metamath Proof Explorer

Theorem elabgt

Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg .) (Contributed by NM, 7-Nov-2005) (Proof shortened by Andrew Salmon, 8-Jun-2011) Reduce axiom usage. (Revised by GG, 12-Oct-2024) (Proof shortened by Wolf Lammen, 11-May-2025) (Proof shortened by SN, 1-Dec-2025)

Ref Expression
Assertion elabgt ( ( 𝐴 ∈ 𝐵 ∧ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) ) → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )

Proof

Step Hyp Ref Expression
1 elab6g ⊢ ( 𝐴 ∈ 𝐵 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ) )
2 pm5.74 ⊢ ( ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) ↔ ( ( 𝑥 = 𝐴 → 𝜑 ) ↔ ( 𝑥 = 𝐴 → 𝜓 ) ) )
3 2 biimpi ⊢ ( ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) → ( ( 𝑥 = 𝐴 → 𝜑 ) ↔ ( 𝑥 = 𝐴 → 𝜓 ) ) )
4 3 alimi ⊢ ( ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) → ∀ 𝑥 ( ( 𝑥 = 𝐴 → 𝜑 ) ↔ ( 𝑥 = 𝐴 → 𝜓 ) ) )
5 albi ⊢ ( ∀ 𝑥 ( ( 𝑥 = 𝐴 → 𝜑 ) ↔ ( 𝑥 = 𝐴 → 𝜓 ) ) → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜓 ) ) )
6 4 5 syl ⊢ ( ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜓 ) ) )
7 1 6 sylan9bb ⊢ ( ( 𝐴 ∈ 𝐵 ∧ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) ) → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜓 ) ) )
8 19.23v ⊢ ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜓 ) ↔ ( ∃ 𝑥 𝑥 = 𝐴 → 𝜓 ) )
9 elisset ⊢ ( 𝐴 ∈ 𝐵 → ∃ 𝑥 𝑥 = 𝐴 )
10 pm5.5 ⊢ ( ∃ 𝑥 𝑥 = 𝐴 → ( ( ∃ 𝑥 𝑥 = 𝐴 → 𝜓 ) ↔ 𝜓 ) )
11 9 10 syl ⊢ ( 𝐴 ∈ 𝐵 → ( ( ∃ 𝑥 𝑥 = 𝐴 → 𝜓 ) ↔ 𝜓 ) )
12 8 11 bitrid ⊢ ( 𝐴 ∈ 𝐵 → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜓 ) ↔ 𝜓 ) )
13 12 adantr ⊢ ( ( 𝐴 ∈ 𝐵 ∧ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) ) → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜓 ) ↔ 𝜓 ) )
14 7 13 bitrd ⊢ ( ( 𝐴 ∈ 𝐵 ∧ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) ) → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )