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Theorem eusvnfb 5329
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 5328 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝑥𝐴)
2 euex 2571 . . . 4 (∃!𝑦𝑥 𝑦 = 𝐴 → ∃𝑦𝑥 𝑦 = 𝐴)
3 eqvisset 3454 . . . . . 6 (𝑦 = 𝐴𝐴 ∈ V)
43sps 2187 . . . . 5 (∀𝑥 𝑦 = 𝐴𝐴 ∈ V)
54exlimiv 1931 . . . 4 (∃𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
62, 5syl 17 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
71, 6jca 511 . 2 (∃!𝑦𝑥 𝑦 = 𝐴 → (𝑥𝐴𝐴 ∈ V))
8 isset 3448 . . . . 5 (𝐴 ∈ V ↔ ∃𝑦 𝑦 = 𝐴)
9 nfcvd 2893 . . . . . . . 8 (𝑥𝐴𝑥𝑦)
10 id 22 . . . . . . . 8 (𝑥𝐴𝑥𝐴)
119, 10nfeqd 2903 . . . . . . 7 (𝑥𝐴 → Ⅎ𝑥 𝑦 = 𝐴)
1211nf5rd 2198 . . . . . 6 (𝑥𝐴 → (𝑦 = 𝐴 → ∀𝑥 𝑦 = 𝐴))
1312eximdv 1918 . . . . 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 5327 . . 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 1539   = wceq 1541  wex 1780  wcel 2110  ∃!weu 2562  wnfc 2877  Vcvv 3434
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2112  ax-9 2120  ax-10 2143  ax-11 2159  ax-12 2179  ax-ext 2702
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-fal 1554  df-ex 1781  df-nf 1785  df-sb 2067  df-mo 2534  df-eu 2563  df-clab 2709  df-cleq 2722  df-clel 2804  df-nfc 2879  df-v 3436  df-sbc 3740  df-csb 3849  df-dif 3903  df-nul 4282
This theorem is referenced by:  eusv2nf  5331  eusv2  5332
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