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Theorem eusvnfb 4448
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 4447 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝑥𝐴)
2 euex 2054 . . . 4 (∃!𝑦𝑥 𝑦 = 𝐴 → ∃𝑦𝑥 𝑦 = 𝐴)
3 id 19 . . . . . . 7 (𝑦 = 𝐴𝑦 = 𝐴)
4 vex 2738 . . . . . . 7 𝑦 ∈ V
53, 4eqeltrrdi 2267 . . . . . 6 (𝑦 = 𝐴𝐴 ∈ V)
65sps 1535 . . . . 5 (∀𝑥 𝑦 = 𝐴𝐴 ∈ V)
76exlimiv 1596 . . . 4 (∃𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
82, 7syl 14 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
91, 8jca 306 . 2 (∃!𝑦𝑥 𝑦 = 𝐴 → (𝑥𝐴𝐴 ∈ V))
10 isset 2741 . . . . 5 (𝐴 ∈ V ↔ ∃𝑦 𝑦 = 𝐴)
11 nfcvd 2318 . . . . . . . 8 (𝑥𝐴𝑥𝑦)
12 id 19 . . . . . . . 8 (𝑥𝐴𝑥𝐴)
1311, 12nfeqd 2332 . . . . . . 7 (𝑥𝐴 → Ⅎ𝑥 𝑦 = 𝐴)
1413nfrd 1518 . . . . . 6 (𝑥𝐴 → (𝑦 = 𝐴 → ∀𝑥 𝑦 = 𝐴))
1514eximdv 1878 . . . . 5 (𝑥𝐴 → (∃𝑦 𝑦 = 𝐴 → ∃𝑦𝑥 𝑦 = 𝐴))
1610, 15biimtrid 152 . . . 4 (𝑥𝐴 → (𝐴 ∈ V → ∃𝑦𝑥 𝑦 = 𝐴))
1716imp 124 . . 3 ((𝑥𝐴𝐴 ∈ V) → ∃𝑦𝑥 𝑦 = 𝐴)
18 eusv1 4446 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴 ↔ ∃𝑦𝑥 𝑦 = 𝐴)
1917, 18sylibr 134 . 2 ((𝑥𝐴𝐴 ∈ V) → ∃!𝑦𝑥 𝑦 = 𝐴)
209, 19impbii 126 1 (∃!𝑦𝑥 𝑦 = 𝐴 ↔ (𝑥𝐴𝐴 ∈ V))
Colors of variables: wff set class
Syntax hints:  wa 104  wb 105  wal 1351   = wceq 1353  wex 1490  ∃!weu 2024  wcel 2146  wnfc 2304  Vcvv 2735
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 709  ax-5 1445  ax-7 1446  ax-gen 1447  ax-ie1 1491  ax-ie2 1492  ax-8 1502  ax-10 1503  ax-11 1504  ax-i12 1505  ax-bndl 1507  ax-4 1508  ax-17 1524  ax-i9 1528  ax-ial 1532  ax-i5r 1533  ax-ext 2157
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1459  df-sb 1761  df-eu 2027  df-mo 2028  df-clab 2162  df-cleq 2168  df-clel 2171  df-nfc 2306  df-v 2737  df-sbc 2961  df-csb 3056
This theorem is referenced by:  eusv2nf  4450  eusv2  4451
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