Theorem eusvnfb 5354

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	⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 ↔ (Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V))

Distinct variable groups:   𝑥,𝑦   𝑦,𝐴

Allowed substitution hint:   𝐴(𝑥)

Proof of Theorem eusvnfb

Step	Hyp	Ref	Expression
1		eusvnf 5353	. . 3 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → Ⅎ𝑥𝐴)
2		euex 2602	. . . 4 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → ∃𝑦∀𝑥 𝑦 = 𝐴)
3		eqvisset 3470	. . . . . 6 ⊢ (𝑦 = 𝐴 → 𝐴 ∈ V)
4	3	sps 2221	. . . . 5 ⊢ (∀𝑥 𝑦 = 𝐴 → 𝐴 ∈ V)
5	4	exlimiv 1962	. . . 4 ⊢ (∃𝑦∀𝑥 𝑦 = 𝐴 → 𝐴 ∈ V)
6	2, 5	syl 18	. . 3 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → 𝐴 ∈ V)
7	1, 6	jca 521	. 2 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → (Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V))
8		isset 3464	. . . . 5 ⊢ (𝐴 ∈ V ↔ ∃𝑦 𝑦 = 𝐴)
9		nfcvd 2923	. . . . . . . 8 ⊢ (Ⅎ𝑥𝐴 → Ⅎ𝑥𝑦)
10		id 23	. . . . . . . 8 ⊢ (Ⅎ𝑥𝐴 → Ⅎ𝑥𝐴)
11	9, 10	nfeqd 2932	. . . . . . 7 ⊢ (Ⅎ𝑥𝐴 → Ⅎ𝑥 𝑦 = 𝐴)
12	11	nf5rd 2232	. . . . . 6 ⊢ (Ⅎ𝑥𝐴 → (𝑦 = 𝐴 → ∀𝑥 𝑦 = 𝐴))
13	12	eximdv 1949	. . . . 5 ⊢ (Ⅎ𝑥𝐴 → (∃𝑦 𝑦 = 𝐴 → ∃𝑦∀𝑥 𝑦 = 𝐴))
14	8, 13	biimtrid 245	. . . 4 ⊢ (Ⅎ𝑥𝐴 → (𝐴 ∈ V → ∃𝑦∀𝑥 𝑦 = 𝐴))
15	14	imp 412	. . 3 ⊢ ((Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V) → ∃𝑦∀𝑥 𝑦 = 𝐴)
16		eusv1 5352	. . 3 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 ↔ ∃𝑦∀𝑥 𝑦 = 𝐴)
17	15, 16	sylibr 237	. 2 ⊢ ((Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V) → ∃!𝑦∀𝑥 𝑦 = 𝐴)
18	7, 17	impbii 212	1 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 ↔ (Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401  ∀wal 1567   = wceq 1569  ∃wex 1811   ∈ wcel 2144  ∃!weu 2593  Ⅎwnfc 2907  Vcvv 3450

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2099  df-mo 2564  df-eu 2594  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-v 3452  df-sbc 3739  df-csb 3847  df-dif 4047  df-nul 4279

This theorem is used by:  eusv2nf  5356  eusv2  5357