Theorem eusvnfb 5399

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 5398	. . 3 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → Ⅎ𝑥𝐴)
2		euex 2575	. . . 4 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → ∃𝑦∀𝑥 𝑦 = 𝐴)
3		eqvisset 3498	. . . . . 6 ⊢ (𝑦 = 𝐴 → 𝐴 ∈ V)
4	3	sps 2183	. . . . 5 ⊢ (∀𝑥 𝑦 = 𝐴 → 𝐴 ∈ V)
5	4	exlimiv 1928	. . . 4 ⊢ (∃𝑦∀𝑥 𝑦 = 𝐴 → 𝐴 ∈ V)
6	2, 5	syl 17	. . 3 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → 𝐴 ∈ V)
7	1, 6	jca 511	. 2 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → (Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V))
8		isset 3492	. . . . 5 ⊢ (𝐴 ∈ V ↔ ∃𝑦 𝑦 = 𝐴)
9		nfcvd 2904	. . . . . . . 8 ⊢ (Ⅎ𝑥𝐴 → Ⅎ𝑥𝑦)
10		id 22	. . . . . . . 8 ⊢ (Ⅎ𝑥𝐴 → Ⅎ𝑥𝐴)
11	9, 10	nfeqd 2914	. . . . . . 7 ⊢ (Ⅎ𝑥𝐴 → Ⅎ𝑥 𝑦 = 𝐴)
12	11	nf5rd 2194	. . . . . 6 ⊢ (Ⅎ𝑥𝐴 → (𝑦 = 𝐴 → ∀𝑥 𝑦 = 𝐴))
13	12	eximdv 1915	. . . . 5 ⊢ (Ⅎ𝑥𝐴 → (∃𝑦 𝑦 = 𝐴 → ∃𝑦∀𝑥 𝑦 = 𝐴))
14	8, 13	biimtrid 242	. . . 4 ⊢ (Ⅎ𝑥𝐴 → (𝐴 ∈ V → ∃𝑦∀𝑥 𝑦 = 𝐴))
15	14	imp 406	. . 3 ⊢ ((Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V) → ∃𝑦∀𝑥 𝑦 = 𝐴)
16		eusv1 5397	. . 3 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 ↔ ∃𝑦∀𝑥 𝑦 = 𝐴)
17	15, 16	sylibr 234	. 2 ⊢ ((Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V) → ∃!𝑦∀𝑥 𝑦 = 𝐴)
18	7, 17	impbii 209	1 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 ↔ (Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V))

Syntax hints:   ↔ wb 206   ∧ wa 395  ∀wal 1535   = wceq 1537  ∃wex 1776   ∈ wcel 2106  ∃!weu 2566  Ⅎwnfc 2888  Vcvv 3478

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 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-10 2139  ax-11 2155  ax-12 2175  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1540  df-fal 1550  df-ex 1777  df-nf 1781  df-sb 2063  df-mo 2538  df-eu 2567  df-clab 2713  df-cleq 2727  df-clel 2814  df-nfc 2890  df-v 3480  df-sbc 3792  df-csb 3909  df-dif 3966  df-nul 4340

This theorem is referenced by:  eusv2nf  5401  eusv2  5402