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Theorem elabgt 2871
Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg 2876.) (Contributed by NM, 7-Nov-2005.) (Proof shortened by Andrew Salmon, 8-Jun-2011.)
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 abid 2158 . . . . . . 7 (𝑥 ∈ {𝑥𝜑} ↔ 𝜑)
2 eleq1 2233 . . . . . . 7 (𝑥 = 𝐴 → (𝑥 ∈ {𝑥𝜑} ↔ 𝐴 ∈ {𝑥𝜑}))
31, 2bitr3id 193 . . . . . 6 (𝑥 = 𝐴 → (𝜑𝐴 ∈ {𝑥𝜑}))
43bibi1d 232 . . . . 5 (𝑥 = 𝐴 → ((𝜑𝜓) ↔ (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
54biimpd 143 . . . 4 (𝑥 = 𝐴 → ((𝜑𝜓) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
65a2i 11 . . 3 ((𝑥 = 𝐴 → (𝜑𝜓)) → (𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
76alimi 1448 . 2 (∀𝑥(𝑥 = 𝐴 → (𝜑𝜓)) → ∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
8 nfcv 2312 . . . 4 𝑥𝐴
9 nfab1 2314 . . . . . 6 𝑥{𝑥𝜑}
109nfel2 2325 . . . . 5 𝑥 𝐴 ∈ {𝑥𝜑}
11 nfv 1521 . . . . 5 𝑥𝜓
1210, 11nfbi 1582 . . . 4 𝑥(𝐴 ∈ {𝑥𝜑} ↔ 𝜓)
13 pm5.5 241 . . . 4 (𝑥 = 𝐴 → ((𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)) ↔ (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
148, 12, 13spcgf 2812 . . 3 (𝐴𝐵 → (∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
1514imp 123 . 2 ((𝐴𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))
167, 15sylan2 284 1 ((𝐴𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑𝜓))) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103  wb 104  wal 1346   = wceq 1348  wcel 2141  {cab 2156
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152
This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732
This theorem is referenced by:  elrab3t  2885
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