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Theorem eusvnfb 5362
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 5361 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝑥𝐴)
2 euex 2604 . . . 4 (∃!𝑦𝑥 𝑦 = 𝐴 → ∃𝑦𝑥 𝑦 = 𝐴)
3 eqvisset 3473 . . . . . 6 (𝑦 = 𝐴𝐴 ∈ V)
43sps 2223 . . . . 5 (∀𝑥 𝑦 = 𝐴𝐴 ∈ V)
54exlimiv 1963 . . . 4 (∃𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
62, 5syl 18 . . 3 (∃!𝑦𝑥 𝑦 = 𝐴𝐴 ∈ V)
71, 6jca 521 . 2 (∃!𝑦𝑥 𝑦 = 𝐴 → (𝑥𝐴𝐴 ∈ V))
8 isset 3467 . . . . 5 (𝐴 ∈ V ↔ ∃𝑦 𝑦 = 𝐴)
9 nfcvd 2925 . . . . . . . 8 (𝑥𝐴𝑥𝑦)
10 id 23 . . . . . . . 8 (𝑥𝐴𝑥𝐴)
119, 10nfeqd 2934 . . . . . . 7 (𝑥𝐴 → Ⅎ𝑥 𝑦 = 𝐴)
1211nf5rd 2234 . . . . . 6 (𝑥𝐴 → (𝑦 = 𝐴 → ∀𝑥 𝑦 = 𝐴))
1312eximdv 1950 . . . . 5 (𝑥𝐴 → (∃𝑦 𝑦 = 𝐴 → ∃𝑦𝑥 𝑦 = 𝐴))
148, 13biimtrid 245 . . . 4 (𝑥𝐴 → (𝐴 ∈ V → ∃𝑦𝑥 𝑦 = 𝐴))
1514imp 412 . . 3 ((𝑥𝐴𝐴 ∈ V) → ∃𝑦𝑥 𝑦 = 𝐴)
16 eusv1 5360 . . 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
This proof depends on syntax axioms:  wb 209  wa 401  wal 1568   = wceq 1570  wex 1812  wcel 2145  ∃!weu 2595  wnfc 2909  Vcvv 3453
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2734
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-mo 2566  df-eu 2596  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-v 3455  df-sbc 3743  df-csb 3851  df-dif 3905  df-nul 4283
This theorem is used by:  eusv2nf  5364  eusv2  5365
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