Theorem vn0OLD 4299

Description: Obsolete version of vn0 4298 as of 12-Jul-2026. (Contributed by NM, 11-Sep-2008.) Avoid ax-8 2145, df-clel 2838. (Revised by GG, 6-Sep-2024.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
vn0OLD	⊢ V ≠ ∅

Proof of Theorem vn0OLD

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		vextru 2748	. . . . . . 7 ⊢ 𝑦 ∈ {𝑥 ∣ ⊤}
2		fal 1584	. . . . . . 7 ⊢ ¬ ⊥
3	1, 2	2th 267	. . . . . 6 ⊢ (𝑦 ∈ {𝑥 ∣ ⊤} ↔ ¬ ⊥)
4		xor3 385	. . . . . 6 ⊢ (¬ (𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥) ↔ (𝑦 ∈ {𝑥 ∣ ⊤} ↔ ¬ ⊥))
5	3, 4	mpbir 234	. . . . 5 ⊢ ¬ (𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥)
6	5	exgen 2004	. . . 4 ⊢ ∃𝑦 ¬ (𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥)
7		exnal 1857	. . . 4 ⊢ (∃𝑦 ¬ (𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥) ↔ ¬ ∀𝑦(𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥))
8	6, 7	mpbi 233	. . 3 ⊢ ¬ ∀𝑦(𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥)
9		dfv2 3458	. . . . 5 ⊢ V = {𝑥 ∣ ⊤}
10		dfnul4 4288	. . . . 5 ⊢ ∅ = {𝑥 ∣ ⊥}
11	9, 10	eqeq12i 2781	. . . 4 ⊢ (V = ∅ ↔ {𝑥 ∣ ⊤} = {𝑥 ∣ ⊥})
12		biidd 265	. . . . 5 ⊢ (𝑥 = 𝑦 → (⊥ ↔ ⊥))
13	12	eqabbw 2836	. . . 4 ⊢ ({𝑥 ∣ ⊤} = {𝑥 ∣ ⊥} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥))
14	11, 13	bitri 278	. . 3 ⊢ (V = ∅ ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥))
15	8, 14	mtbir 326	. 2 ⊢ ¬ V = ∅
16	15	neir 2961	1 ⊢ V ≠ ∅

Syntax hints:  ¬ wn 3   ↔ wb 209  ∀wal 1568   = wceq 1570  ⊤wtru 1571  ⊥wfal 1582  ∃wex 1809   ∈ wcel 2143  {cab 2741   ≠ wne 2958  Vcvv 3455  ∅c0 4286

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-ne 2959  df-v 3457  df-dif 3908  df-nul 4287

This theorem is referenced by: (None)