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Theorem elabgt 3628
Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg 3633.) (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 3625 . . 3 (𝐴𝐵 → (𝐴 ∈ {𝑥𝜑} ↔ ∀𝑥(𝑥 = 𝐴𝜑)))
2 pm5.74 270 . . . . . 6 ((𝑥 = 𝐴 → (𝜑𝜓)) ↔ ((𝑥 = 𝐴𝜑) ↔ (𝑥 = 𝐴𝜓)))
32biimpi 216 . . . . 5 ((𝑥 = 𝐴 → (𝜑𝜓)) → ((𝑥 = 𝐴𝜑) ↔ (𝑥 = 𝐴𝜓)))
43alimi 1813 . . . 4 (∀𝑥(𝑥 = 𝐴 → (𝜑𝜓)) → ∀𝑥((𝑥 = 𝐴𝜑) ↔ (𝑥 = 𝐴𝜓)))
5 albi 1820 . . . 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 509 . 2 ((𝐴𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑𝜓))) → (𝐴 ∈ {𝑥𝜑} ↔ ∀𝑥(𝑥 = 𝐴𝜓)))
8 19.23v 1944 . . . 4 (∀𝑥(𝑥 = 𝐴𝜓) ↔ (∃𝑥 𝑥 = 𝐴𝜓))
9 elisset 2819 . . . . 5 (𝐴𝐵 → ∃𝑥 𝑥 = 𝐴)
10 pm5.5 361 . . . . 5 (∃𝑥 𝑥 = 𝐴 → ((∃𝑥 𝑥 = 𝐴𝜓) ↔ 𝜓))
119, 10syl 17 . . . 4 (𝐴𝐵 → ((∃𝑥 𝑥 = 𝐴𝜓) ↔ 𝜓))
128, 11bitrid 283 . . 3 (𝐴𝐵 → (∀𝑥(𝑥 = 𝐴𝜓) ↔ 𝜓))
1312adantr 480 . 2 ((𝐴𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑𝜓))) → (∀𝑥(𝑥 = 𝐴𝜓) ↔ 𝜓))
147, 13bitrd 279 1 ((𝐴𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑𝜓))) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395  wal 1540   = wceq 1542  wex 1781  wcel 2114  {cab 2715
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709
This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1545  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812
This theorem is referenced by:  elabg  3633  elrab3t  3647  dfrtrcl2  14997  iinabrex  32655  abfmpeld  32743  abfmpel  32744  bj-elgab  37181  dftrcl3  44070  dfrtrcl3  44083
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