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Theorem eusvnfb 5393
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 5392 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝑥𝐴)
2 euex 2577 . . . 4 (∃!𝑦𝑥 𝑦 = 𝐴 → ∃𝑦𝑥 𝑦 = 𝐴)
3 eqvisset 3500 . . . . . 6 (𝑦 = 𝐴𝐴 ∈ V)
43sps 2185 . . . . 5 (∀𝑥 𝑦 = 𝐴𝐴 ∈ V)
54exlimiv 1930 . . . 4 (∃𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
62, 5syl 17 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
71, 6jca 511 . 2 (∃!𝑦𝑥 𝑦 = 𝐴 → (𝑥𝐴𝐴 ∈ V))
8 isset 3494 . . . . 5 (𝐴 ∈ V ↔ ∃𝑦 𝑦 = 𝐴)
9 nfcvd 2906 . . . . . . . 8 (𝑥𝐴𝑥𝑦)
10 id 22 . . . . . . . 8 (𝑥𝐴𝑥𝐴)
119, 10nfeqd 2916 . . . . . . 7 (𝑥𝐴 → Ⅎ𝑥 𝑦 = 𝐴)
1211nf5rd 2196 . . . . . 6 (𝑥𝐴 → (𝑦 = 𝐴 → ∀𝑥 𝑦 = 𝐴))
1312eximdv 1917 . . . . 5 (𝑥𝐴 → (∃𝑦 𝑦 = 𝐴 → ∃𝑦𝑥 𝑦 = 𝐴))
148, 13biimtrid 242 . . . 4 (𝑥𝐴 → (𝐴 ∈ V → ∃𝑦𝑥 𝑦 = 𝐴))
1514imp 406 . . 3 ((𝑥𝐴𝐴 ∈ V) → ∃𝑦𝑥 𝑦 = 𝐴)
16 eusv1 5391 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴 ↔ ∃𝑦𝑥 𝑦 = 𝐴)
1715, 16sylibr 234 . 2 ((𝑥𝐴𝐴 ∈ V) → ∃!𝑦𝑥 𝑦 = 𝐴)
187, 17impbii 209 1 (∃!𝑦𝑥 𝑦 = 𝐴 ↔ (𝑥𝐴𝐴 ∈ V))
Colors of variables: wff setvar class
Syntax hints:  wb 206  wa 395  wal 1538   = wceq 1540  wex 1779  wcel 2108  ∃!weu 2568  wnfc 2890  Vcvv 3480
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2065  df-mo 2540  df-eu 2569  df-clab 2715  df-cleq 2729  df-clel 2816  df-nfc 2892  df-v 3482  df-sbc 3789  df-csb 3900  df-dif 3954  df-nul 4334
This theorem is referenced by:  eusv2nf  5395  eusv2  5396
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