Theorem eusvnfb 5349

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 5348	. . 3 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → Ⅎ𝑥𝐴)
2		euex 2603	. . . 4 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → ∃𝑦∀𝑥 𝑦 = 𝐴)
3		eqvisset 3473	. . . . . 6 ⊢ (𝑦 = 𝐴 → 𝐴 ∈ V)
4	3	sps 2219	. . . . 5 ⊢ (∀𝑥 𝑦 = 𝐴 → 𝐴 ∈ V)
5	4	exlimiv 1949	. . . 4 ⊢ (∃𝑦∀𝑥 𝑦 = 𝐴 → 𝐴 ∈ V)
6	2, 5	syl 17	. . 3 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → 𝐴 ∈ V)
7	1, 6	jca 519	. 2 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 → (Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V))
8		isset 3467	. . . . 5 ⊢ (𝐴 ∈ V ↔ ∃𝑦 𝑦 = 𝐴)
9		nfcvd 2924	. . . . . . . 8 ⊢ (Ⅎ𝑥𝐴 → Ⅎ𝑥𝑦)
10		id 22	. . . . . . . 8 ⊢ (Ⅎ𝑥𝐴 → Ⅎ𝑥𝐴)
11	9, 10	nfeqd 2933	. . . . . . 7 ⊢ (Ⅎ𝑥𝐴 → Ⅎ𝑥 𝑦 = 𝐴)
12	11	nf5rd 2230	. . . . . 6 ⊢ (Ⅎ𝑥𝐴 → (𝑦 = 𝐴 → ∀𝑥 𝑦 = 𝐴))
13	12	eximdv 1936	. . . . 5 ⊢ (Ⅎ𝑥𝐴 → (∃𝑦 𝑦 = 𝐴 → ∃𝑦∀𝑥 𝑦 = 𝐴))
14	8, 13	biimtrid 244	. . . 4 ⊢ (Ⅎ𝑥𝐴 → (𝐴 ∈ V → ∃𝑦∀𝑥 𝑦 = 𝐴))
15	14	imp 410	. . 3 ⊢ ((Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V) → ∃𝑦∀𝑥 𝑦 = 𝐴)
16		eusv1 5347	. . 3 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 ↔ ∃𝑦∀𝑥 𝑦 = 𝐴)
17	15, 16	sylibr 236	. 2 ⊢ ((Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V) → ∃!𝑦∀𝑥 𝑦 = 𝐴)
18	7, 17	impbii 211	1 ⊢ (∃!𝑦∀𝑥 𝑦 = 𝐴 ↔ (Ⅎ𝑥𝐴 ∧ 𝐴 ∈ V))

Syntax hints:   ↔ wb 208   ∧ wa 399  ∀wal 1557   = wceq 1559  ∃wex 1798   ∈ wcel 2141  ∃!weu 2594  Ⅎwnfc 2908  Vcvv 3453

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-10 2174  ax-11 2190  ax-12 2211  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1562  df-fal 1572  df-ex 1799  df-nf 1803  df-sb 2090  df-mo 2565  df-eu 2595  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-v 3455  df-sbc 3745  df-csb 3853  df-dif 3907  df-nul 4286

This theorem is referenced by:  eusv2nf  5351  eusv2  5352