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Theorem eusvnfb 4501
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 4500 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝑥𝐴)
2 euex 2084 . . . 4 (∃!𝑦𝑥 𝑦 = 𝐴 → ∃𝑦𝑥 𝑦 = 𝐴)
3 id 19 . . . . . . 7 (𝑦 = 𝐴𝑦 = 𝐴)
4 vex 2775 . . . . . . 7 𝑦 ∈ V
53, 4eqeltrrdi 2297 . . . . . 6 (𝑦 = 𝐴𝐴 ∈ V)
65sps 1560 . . . . 5 (∀𝑥 𝑦 = 𝐴𝐴 ∈ V)
76exlimiv 1621 . . . 4 (∃𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
82, 7syl 14 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
91, 8jca 306 . 2 (∃!𝑦𝑥 𝑦 = 𝐴 → (𝑥𝐴𝐴 ∈ V))
10 isset 2778 . . . . 5 (𝐴 ∈ V ↔ ∃𝑦 𝑦 = 𝐴)
11 nfcvd 2349 . . . . . . . 8 (𝑥𝐴𝑥𝑦)
12 id 19 . . . . . . . 8 (𝑥𝐴𝑥𝐴)
1311, 12nfeqd 2363 . . . . . . 7 (𝑥𝐴 → Ⅎ𝑥 𝑦 = 𝐴)
1413nfrd 1543 . . . . . 6 (𝑥𝐴 → (𝑦 = 𝐴 → ∀𝑥 𝑦 = 𝐴))
1514eximdv 1903 . . . . 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 4499 . . 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 1371   = wceq 1373  wex 1515  ∃!weu 2054  wcel 2176  wnfc 2335  Vcvv 2772
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 711  ax-5 1470  ax-7 1471  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-10 1528  ax-11 1529  ax-i12 1530  ax-bndl 1532  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-i5r 1558  ax-ext 2187
This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1484  df-sb 1786  df-eu 2057  df-mo 2058  df-clab 2192  df-cleq 2198  df-clel 2201  df-nfc 2337  df-v 2774  df-sbc 2999  df-csb 3094
This theorem is referenced by:  eusv2nf  4503  eusv2  4504
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