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Theorem eusvnfb 5259
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 5258 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝑥𝐴)
2 euex 2637 . . . 4 (∃!𝑦𝑥 𝑦 = 𝐴 → ∃𝑦𝑥 𝑦 = 𝐴)
3 eqvisset 3458 . . . . . 6 (𝑦 = 𝐴𝐴 ∈ V)
43sps 2182 . . . . 5 (∀𝑥 𝑦 = 𝐴𝐴 ∈ V)
54exlimiv 1931 . . . 4 (∃𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
62, 5syl 17 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
71, 6jca 515 . 2 (∃!𝑦𝑥 𝑦 = 𝐴 → (𝑥𝐴𝐴 ∈ V))
8 isset 3453 . . . . 5 (𝐴 ∈ V ↔ ∃𝑦 𝑦 = 𝐴)
9 nfcvd 2956 . . . . . . . 8 (𝑥𝐴𝑥𝑦)
10 id 22 . . . . . . . 8 (𝑥𝐴𝑥𝐴)
119, 10nfeqd 2965 . . . . . . 7 (𝑥𝐴 → Ⅎ𝑥 𝑦 = 𝐴)
1211nf5rd 2194 . . . . . 6 (𝑥𝐴 → (𝑦 = 𝐴 → ∀𝑥 𝑦 = 𝐴))
1312eximdv 1918 . . . . 5 (𝑥𝐴 → (∃𝑦 𝑦 = 𝐴 → ∃𝑦𝑥 𝑦 = 𝐴))
148, 13syl5bi 245 . . . 4 (𝑥𝐴 → (𝐴 ∈ V → ∃𝑦𝑥 𝑦 = 𝐴))
1514imp 410 . . 3 ((𝑥𝐴𝐴 ∈ V) → ∃𝑦𝑥 𝑦 = 𝐴)
16 eusv1 5257 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴 ↔ ∃𝑦𝑥 𝑦 = 𝐴)
1715, 16sylibr 237 . 2 ((𝑥𝐴𝐴 ∈ V) → ∃!𝑦𝑥 𝑦 = 𝐴)
187, 17impbii 212 1 (∃!𝑦𝑥 𝑦 = 𝐴 ↔ (𝑥𝐴𝐴 ∈ V))
Colors of variables: wff setvar class
Syntax hints:  wb 209  wa 399  wal 1536   = wceq 1538  wex 1781  wcel 2111  ∃!weu 2628  wnfc 2936  Vcvv 3441
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-10 2142  ax-11 2158  ax-12 2175  ax-ext 2770
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-fal 1551  df-ex 1782  df-nf 1786  df-sb 2070  df-mo 2598  df-eu 2629  df-clab 2777  df-cleq 2791  df-clel 2870  df-nfc 2938  df-v 3443  df-sbc 3721  df-csb 3829  df-dif 3884  df-nul 4244
This theorem is referenced by:  eusv2nf  5261  eusv2  5262
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