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Theorem eusvnfb 4186
Description: Two ways to say that 𝐴(𝑥) is a set expression that does not depend on 𝑥. (Contributed by Mario Carneiro, 18-Nov-2016.)
Assertion
Ref Expression
eusvnfb (∃!𝑦𝑥 𝑦 = 𝐴 ↔ (𝑥𝐴𝐴 ∈ V))
Distinct variable groups:   𝑥,𝑦   𝑦,𝐴
Allowed substitution hint:   𝐴(𝑥)

Proof of Theorem eusvnfb
StepHypRef Expression
1 eusvnf 4185 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝑥𝐴)
2 euex 1930 . . . 4 (∃!𝑦𝑥 𝑦 = 𝐴 → ∃𝑦𝑥 𝑦 = 𝐴)
3 id 19 . . . . . . 7 (𝑦 = 𝐴𝑦 = 𝐴)
4 vex 2560 . . . . . . 7 𝑦 ∈ V
53, 4syl6eqelr 2129 . . . . . 6 (𝑦 = 𝐴𝐴 ∈ V)
65sps 1430 . . . . 5 (∀𝑥 𝑦 = 𝐴𝐴 ∈ V)
76exlimiv 1489 . . . 4 (∃𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
82, 7syl 14 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
91, 8jca 290 . 2 (∃!𝑦𝑥 𝑦 = 𝐴 → (𝑥𝐴𝐴 ∈ V))
10 isset 2561 . . . . 5 (𝐴 ∈ V ↔ ∃𝑦 𝑦 = 𝐴)
11 nfcvd 2179 . . . . . . . 8 (𝑥𝐴𝑥𝑦)
12 id 19 . . . . . . . 8 (𝑥𝐴𝑥𝐴)
1311, 12nfeqd 2192 . . . . . . 7 (𝑥𝐴 → Ⅎ𝑥 𝑦 = 𝐴)
1413nfrd 1413 . . . . . 6 (𝑥𝐴 → (𝑦 = 𝐴 → ∀𝑥 𝑦 = 𝐴))
1514eximdv 1760 . . . . 5 (𝑥𝐴 → (∃𝑦 𝑦 = 𝐴 → ∃𝑦𝑥 𝑦 = 𝐴))
1610, 15syl5bi 141 . . . 4 (𝑥𝐴 → (𝐴 ∈ V → ∃𝑦𝑥 𝑦 = 𝐴))
1716imp 115 . . 3 ((𝑥𝐴𝐴 ∈ V) → ∃𝑦𝑥 𝑦 = 𝐴)
18 eusv1 4184 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴 ↔ ∃𝑦𝑥 𝑦 = 𝐴)
1917, 18sylibr 137 . 2 ((𝑥𝐴𝐴 ∈ V) → ∃!𝑦𝑥 𝑦 = 𝐴)
209, 19impbii 117 1 (∃!𝑦𝑥 𝑦 = 𝐴 ↔ (𝑥𝐴𝐴 ∈ V))
Colors of variables: wff set class
Syntax hints:  wa 97  wb 98  wal 1241   = wceq 1243  wex 1381  wcel 1393  ∃!weu 1900  wnfc 2165  Vcvv 2557
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022
This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-eu 1903  df-mo 1904  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559  df-sbc 2765  df-csb 2853
This theorem is referenced by:  eusv2nf  4188  eusv2  4189
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