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Theorem elabgt 2967
Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg 2972.) (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 2226 . . . . . . 7 (𝑥 ∈ {𝑥𝜑} ↔ 𝜑)
2 eleq1 2301 . . . . . . 7 (𝑥 = 𝐴 → (𝑥 ∈ {𝑥𝜑} ↔ 𝐴 ∈ {𝑥𝜑}))
31, 2bitr3id 194 . . . . . 6 (𝑥 = 𝐴 → (𝜑𝐴 ∈ {𝑥𝜑}))
43bibi1d 233 . . . . 5 (𝑥 = 𝐴 → ((𝜑𝜓) ↔ (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
54biimpd 144 . . . 4 (𝑥 = 𝐴 → ((𝜑𝜓) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
65a2i 11 . . 3 ((𝑥 = 𝐴 → (𝜑𝜓)) → (𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
76alimi 1508 . 2 (∀𝑥(𝑥 = 𝐴 → (𝜑𝜓)) → ∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
8 nfcv 2392 . . . 4 𝑥𝐴
9 nfab1 2394 . . . . . 6 𝑥{𝑥𝜑}
109nfel2 2405 . . . . 5 𝑥 𝐴 ∈ {𝑥𝜑}
11 nfv 1581 . . . . 5 𝑥𝜓
1210, 11nfbi 1642 . . . 4 𝑥(𝐴 ∈ {𝑥𝜑} ↔ 𝜓)
13 pm5.5 242 . . . 4 (𝑥 = 𝐴 → ((𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)) ↔ (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
148, 12, 13spcgf 2907 . . 3 (𝐴𝐵 → (∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
1514imp 124 . 2 ((𝐴𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))
167, 15sylan2 286 1 ((𝐴𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑𝜓))) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104  wb 105  wal 1400   = wceq 1402  wcel 2209  {cab 2224
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220
This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823
This theorem is referenced by:  elrab3t  2981
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