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Theorem elabgt 2946
Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg 2951.) (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 2218 . . . . . . 7 (𝑥 ∈ {𝑥𝜑} ↔ 𝜑)
2 eleq1 2293 . . . . . . 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 1503 . 2 (∀𝑥(𝑥 = 𝐴 → (𝜑𝜓)) → ∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
8 nfcv 2373 . . . 4 𝑥𝐴
9 nfab1 2375 . . . . . 6 𝑥{𝑥𝜑}
109nfel2 2386 . . . . 5 𝑥 𝐴 ∈ {𝑥𝜑}
11 nfv 1576 . . . . 5 𝑥𝜓
1210, 11nfbi 1637 . . . 4 𝑥(𝐴 ∈ {𝑥𝜑} ↔ 𝜓)
13 pm5.5 242 . . . 4 (𝑥 = 𝐴 → ((𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)) ↔ (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
148, 12, 13spcgf 2887 . . 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 1395   = wceq 1397  wcel 2201  {cab 2216
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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2212
This theorem depends on definitions:  df-bi 117  df-tru 1400  df-nf 1509  df-sb 1810  df-clab 2217  df-cleq 2223  df-clel 2226  df-nfc 2362  df-v 2803
This theorem is referenced by:  elrab3t  2960
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