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Theorem eusvnfb 5285
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 5284 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝑥𝐴)
2 euex 2658 . . . 4 (∃!𝑦𝑥 𝑦 = 𝐴 → ∃𝑦𝑥 𝑦 = 𝐴)
3 eqvisset 3511 . . . . . 6 (𝑦 = 𝐴𝐴 ∈ V)
43sps 2180 . . . . 5 (∀𝑥 𝑦 = 𝐴𝐴 ∈ V)
54exlimiv 1927 . . . 4 (∃𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
62, 5syl 17 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
71, 6jca 514 . 2 (∃!𝑦𝑥 𝑦 = 𝐴 → (𝑥𝐴𝐴 ∈ V))
8 isset 3506 . . . . 5 (𝐴 ∈ V ↔ ∃𝑦 𝑦 = 𝐴)
9 nfcvd 2978 . . . . . . . 8 (𝑥𝐴𝑥𝑦)
10 id 22 . . . . . . . 8 (𝑥𝐴𝑥𝐴)
119, 10nfeqd 2988 . . . . . . 7 (𝑥𝐴 → Ⅎ𝑥 𝑦 = 𝐴)
1211nf5rd 2192 . . . . . 6 (𝑥𝐴 → (𝑦 = 𝐴 → ∀𝑥 𝑦 = 𝐴))
1312eximdv 1914 . . . . 5 (𝑥𝐴 → (∃𝑦 𝑦 = 𝐴 → ∃𝑦𝑥 𝑦 = 𝐴))
148, 13syl5bi 244 . . . 4 (𝑥𝐴 → (𝐴 ∈ V → ∃𝑦𝑥 𝑦 = 𝐴))
1514imp 409 . . 3 ((𝑥𝐴𝐴 ∈ V) → ∃𝑦𝑥 𝑦 = 𝐴)
16 eusv1 5283 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴 ↔ ∃𝑦𝑥 𝑦 = 𝐴)
1715, 16sylibr 236 . 2 ((𝑥𝐴𝐴 ∈ V) → ∃!𝑦𝑥 𝑦 = 𝐴)
187, 17impbii 211 1 (∃!𝑦𝑥 𝑦 = 𝐴 ↔ (𝑥𝐴𝐴 ∈ V))
Colors of variables: wff setvar class
Syntax hints:  wb 208  wa 398  wal 1531   = wceq 1533  wex 1776  wcel 2110  ∃!weu 2649  wnfc 2961  Vcvv 3494
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1536  df-fal 1546  df-ex 1777  df-nf 1781  df-sb 2066  df-mo 2618  df-eu 2650  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-v 3496  df-sbc 3772  df-csb 3883  df-dif 3938  df-nul 4291
This theorem is referenced by:  eusv2nf  5287  eusv2  5288
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