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Theorem elabgt 2707
Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg 2711.) (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 2044 . . . . . . 7 (𝑥 ∈ {𝑥𝜑} ↔ 𝜑)
2 eleq1 2116 . . . . . . 7 (𝑥 = 𝐴 → (𝑥 ∈ {𝑥𝜑} ↔ 𝐴 ∈ {𝑥𝜑}))
31, 2syl5bbr 187 . . . . . 6 (𝑥 = 𝐴 → (𝜑𝐴 ∈ {𝑥𝜑}))
43bibi1d 226 . . . . 5 (𝑥 = 𝐴 → ((𝜑𝜓) ↔ (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
54biimpd 136 . . . 4 (𝑥 = 𝐴 → ((𝜑𝜓) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
65a2i 11 . . 3 ((𝑥 = 𝐴 → (𝜑𝜓)) → (𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
76alimi 1360 . 2 (∀𝑥(𝑥 = 𝐴 → (𝜑𝜓)) → ∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
8 nfcv 2194 . . . 4 𝑥𝐴
9 nfab1 2196 . . . . . 6 𝑥{𝑥𝜑}
109nfel2 2206 . . . . 5 𝑥 𝐴 ∈ {𝑥𝜑}
11 nfv 1437 . . . . 5 𝑥𝜓
1210, 11nfbi 1497 . . . 4 𝑥(𝐴 ∈ {𝑥𝜑} ↔ 𝜓)
13 pm5.5 235 . . . 4 (𝑥 = 𝐴 → ((𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)) ↔ (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
148, 12, 13spcgf 2652 . . 3 (𝐴𝐵 → (∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)))
1514imp 119 . 2 ((𝐴𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))
167, 15sylan2 274 1 ((𝐴𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑𝜓))) → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 101  wb 102  wal 1257   = wceq 1259  wcel 1409  {cab 2042
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038
This theorem depends on definitions:  df-bi 114  df-tru 1262  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-nfc 2183  df-v 2576
This theorem is referenced by:  elrab3t  2720
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