Theorem ab0 3569

Description: Empty class abstraction. (Contributed by SF, 5-Jan-2018.)

Assertion

Ref	Expression
ab0	⊢ ({x ∣ φ} = ∅ ↔ ∀x ¬ φ)

Proof of Theorem ab0

Step	Hyp	Ref	Expression
1		abn0 3568	. . 3 ⊢ ({x ∣ φ} ≠ ∅ ↔ ∃xφ)
2		df-ne 2518	. . 3 ⊢ ({x ∣ φ} ≠ ∅ ↔ ¬ {x ∣ φ} = ∅)
3		df-ex 1542	. . 3 ⊢ (∃xφ ↔ ¬ ∀x ¬ φ)
4	1, 2, 3	3bitr3i 266	. 2 ⊢ (¬ {x ∣ φ} = ∅ ↔ ¬ ∀x ¬ φ)
5	4	con4bii 288	1 ⊢ ({x ∣ φ} = ∅ ↔ ∀x ¬ φ)

Syntax hints:  ¬ wn 3   ↔ wb 176  ∀wal 1540  ∃wex 1541   = wceq 1642  {cab 2339   ≠ wne 2516  ∅c0 3550

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-ne 2518  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-dif 3215  df-nul 3551

This theorem is referenced by:  setswith  4321  nnadjoin  4520  tfinnn  4534