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Theorem elabgt 3617
Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg 3621.) (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.)
Assertion
Ref Expression
elabgt ((𝐴𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑𝜓))) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))
Distinct variable groups:   𝑥,𝐴   𝜓,𝑥
Allowed substitution hints:   𝜑(𝑥)   𝐵(𝑥)

Proof of Theorem elabgt
StepHypRef Expression
1 elab6g 3614 . . 3 (𝐴𝐵 → (𝐴 ∈ {𝑥𝜑} ↔ ∀𝑥(𝑥 = 𝐴𝜑)))
2 pm5.74 271 . . . . . 6 ((𝑥 = 𝐴 → (𝜑𝜓)) ↔ ((𝑥 = 𝐴𝜑) ↔ (𝑥 = 𝐴𝜓)))
32biimpi 217 . . . . 5 ((𝑥 = 𝐴 → (𝜑𝜓)) → ((𝑥 = 𝐴𝜑) ↔ (𝑥 = 𝐴𝜓)))
43alimi 1818 . . . 4 (∀𝑥(𝑥 = 𝐴 → (𝜑𝜓)) → ∀𝑥((𝑥 = 𝐴𝜑) ↔ (𝑥 = 𝐴𝜓)))
5 albi 1825 . . . 4 (∀𝑥((𝑥 = 𝐴𝜑) ↔ (𝑥 = 𝐴𝜓)) → (∀𝑥(𝑥 = 𝐴𝜑) ↔ ∀𝑥(𝑥 = 𝐴𝜓)))
64, 5syl 17 . . 3 (∀𝑥(𝑥 = 𝐴 → (𝜑𝜓)) → (∀𝑥(𝑥 = 𝐴𝜑) ↔ ∀𝑥(𝑥 = 𝐴𝜓)))
71, 6sylan9bb 514 . 2 ((𝐴𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑𝜓))) → (𝐴 ∈ {𝑥𝜑} ↔ ∀𝑥(𝑥 = 𝐴𝜓)))
8 19.23v 1949 . . . 4 (∀𝑥(𝑥 = 𝐴𝜓) ↔ (∃𝑥 𝑥 = 𝐴𝜓))
9 elisset 2822 . . . . 5 (𝐴𝐵 → ∃𝑥 𝑥 = 𝐴)
10 pm5.5 362 . . . . 5 (∃𝑥 𝑥 = 𝐴 → ((∃𝑥 𝑥 = 𝐴𝜓) ↔ 𝜓))
119, 10syl 17 . . . 4 (𝐴𝐵 → ((∃𝑥 𝑥 = 𝐴𝜓) ↔ 𝜓))
128, 11bitrid 284 . . 3 (𝐴𝐵 → (∀𝑥(𝑥 = 𝐴𝜓) ↔ 𝜓))
1312adantr 481 . 2 ((𝐴𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑𝜓))) → (∀𝑥(𝑥 = 𝐴𝜓) ↔ 𝜓))
147, 13bitrd 280 1 ((𝐴𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑𝜓))) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 207  wa 396  wal 1545   = wceq 1547  wex 1786  wcel 2119  {cab 2718
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2712
This theorem depends on definitions:  df-bi 208  df-an 397  df-tru 1550  df-ex 1787  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815
This theorem is referenced by:  elabg  3621  elrab3t  3635  dfrtrcl2  15022  iinabrex  32665  abfmpeld  32753  abfmpel  32754  bj-elgab  37299  dftrcl3  44171  dfrtrcl3  44184
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