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Theorem elabgt 2893
Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg 2898.) (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 2177 . . . . . . 7 (𝑥 ∈ {𝑥𝜑} ↔ 𝜑)
2 eleq1 2252 . . . . . . 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 1466 . 2 (∀𝑥(𝑥 = 𝐴 → (𝜑𝜓)) → ∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
8 nfcv 2332 . . . 4 𝑥𝐴
9 nfab1 2334 . . . . . 6 𝑥{𝑥𝜑}
109nfel2 2345 . . . . 5 𝑥 𝐴 ∈ {𝑥𝜑}
11 nfv 1539 . . . . 5 𝑥𝜓
1210, 11nfbi 1600 . . . 4 𝑥(𝐴 ∈ {𝑥𝜑} ↔ 𝜓)
13 pm5.5 242 . . . 4 (𝑥 = 𝐴 → ((𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)) ↔ (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
148, 12, 13spcgf 2834 . . 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 1362   = wceq 1364  wcel 2160  {cab 2175
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 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2171
This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2176  df-cleq 2182  df-clel 2185  df-nfc 2321  df-v 2754
This theorem is referenced by:  elrab3t  2907
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