Theorem eusvnfb 5390

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 5389	. . 3 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → Ⅎ𝑥𝐴)
2		euex 2569	. . . 4 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → ∃𝑦∀𝑥 𝑦 = 𝐴)
3		eqvisset 3490	. . . . . 6 ⊢ (𝑦 = 𝐴 → 𝐴 ∈ V)
4	3	sps 2176	. . . . 5 ⊢ (∀𝑥 𝑦 = 𝐴 → 𝐴 ∈ V)
5	4	exlimiv 1931	. . . 4 ⊢ (∃𝑦∀𝑥 𝑦 = 𝐴 → 𝐴 ∈ V)
6	2, 5	syl 17	. . 3 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → 𝐴 ∈ V)
7	1, 6	jca 510	. 2 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → (Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V))
8		isset 3485	. . . . 5 ⊢ (𝐴 ∈ V ↔ ∃𝑦 𝑦 = 𝐴)
9		nfcvd 2902	. . . . . . . 8 ⊢ (Ⅎ𝑥𝐴 → Ⅎ𝑥𝑦)
10		id 22	. . . . . . . 8 ⊢ (Ⅎ𝑥𝐴 → Ⅎ𝑥𝐴)
11	9, 10	nfeqd 2911	. . . . . . 7 ⊢ (Ⅎ𝑥𝐴 → Ⅎ𝑥 𝑦 = 𝐴)
12	11	nf5rd 2187	. . . . . 6 ⊢ (Ⅎ𝑥𝐴 → (𝑦 = 𝐴 → ∀𝑥 𝑦 = 𝐴))
13	12	eximdv 1918	. . . . 5 ⊢ (Ⅎ𝑥𝐴 → (∃𝑦 𝑦 = 𝐴 → ∃𝑦∀𝑥 𝑦 = 𝐴))
14	8, 13	biimtrid 241	. . . 4 ⊢ (Ⅎ𝑥𝐴 → (𝐴 ∈ V → ∃𝑦∀𝑥 𝑦 = 𝐴))
15	14	imp 405	. . 3 ⊢ ((Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V) → ∃𝑦∀𝑥 𝑦 = 𝐴)
16		eusv1 5388	. . 3 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 ↔ ∃𝑦∀𝑥 𝑦 = 𝐴)
17	15, 16	sylibr 233	. 2 ⊢ ((Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V) → ∃!𝑦∀𝑥 𝑦 = 𝐴)
18	7, 17	impbii 208	1 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 ↔ (Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V))

Syntax hints:   ↔ wb 205   ∧ wa 394  ∀wal 1537   = wceq 1539  ∃wex 1779   ∈ wcel 2104  ∃!weu 2560  Ⅎwnfc 2881  Vcvv 3472

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-10 2135  ax-11 2152  ax-12 2169  ax-ext 2701

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 844  df-tru 1542  df-fal 1552  df-ex 1780  df-nf 1784  df-sb 2066  df-mo 2532  df-eu 2561  df-clab 2708  df-cleq 2722  df-clel 2808  df-nfc 2883  df-v 3474  df-sbc 3777  df-csb 3893  df-dif 3950  df-nul 4322

This theorem is referenced by:  eusv2nf  5392  eusv2  5393