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Theorem vn0OLD 4307
Description: Obsolete version of vn0 4306 as of 12-Jul-2026. (Contributed by NM, 11-Sep-2008.) Avoid ax-8 2151, df-clel 2844. (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.
StepHypRef Expression
1 vextru 2754 . . . . . . 7 𝑦 ∈ {𝑥 ∣ ⊤}
2 fal 1581 . . . . . . 7 ¬ ⊥
31, 22th 267 . . . . . 6 (𝑦 ∈ {𝑥 ∣ ⊤} ↔ ¬ ⊥)
4 xor3 385 . . . . . 6 (¬ (𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥) ↔ (𝑦 ∈ {𝑥 ∣ ⊤} ↔ ¬ ⊥))
53, 4mpbir 234 . . . . 5 ¬ (𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥)
65exgen 2001 . . . 4 𝑦 ¬ (𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥)
7 exnal 1854 . . . 4 (∃𝑦 ¬ (𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥) ↔ ¬ ∀𝑦(𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥))
86, 7mpbi 233 . . 3 ¬ ∀𝑦(𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥)
9 dfv2 3466 . . . . 5 V = {𝑥 ∣ ⊤}
10 dfnul4 4296 . . . . 5 ∅ = {𝑥 ∣ ⊥}
119, 10eqeq12i 2787 . . . 4 (V = ∅ ↔ {𝑥 ∣ ⊤} = {𝑥 ∣ ⊥})
12 biidd 265 . . . . 5 (𝑥 = 𝑦 → (⊥ ↔ ⊥))
1312eqabbw 2842 . . . 4 ({𝑥 ∣ ⊤} = {𝑥 ∣ ⊥} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥))
1411, 13bitri 278 . . 3 (V = ∅ ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ ⊤} ↔ ⊥))
158, 14mtbir 326 . 2 ¬ V = ∅
1615neir 2967 1 V ≠ ∅
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 209  wal 1565   = wceq 1567  wtru 1568  wfal 1579  wex 1806  wcel 2149  {cab 2747  wne 2964  Vcvv 3463  c0 4294
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-9 2159  ax-ext 2741
This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-ne 2965  df-v 3465  df-dif 3916  df-nul 4295
This theorem is referenced by: (None)
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