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Theorem ralfal 44404
Description: Two ways of expressing empty set. (Contributed by Glauco Siliprandi, 24-Jan-2024.)
Hypothesis
Ref Expression
ralfal.1 𝑥𝐴
Assertion
Ref Expression
ralfal (𝐴 = ∅ ↔ ∀𝑥𝐴 ⊥)

Proof of Theorem ralfal
StepHypRef Expression
1 df-fal 1546 . . . 4 (⊥ ↔ ¬ ⊤)
21ralbii 3085 . . 3 (∀𝑥𝐴 ⊥ ↔ ∀𝑥𝐴 ¬ ⊤)
3 ralnex 3064 . . 3 (∀𝑥𝐴 ¬ ⊤ ↔ ¬ ∃𝑥𝐴 ⊤)
42, 3bitri 275 . 2 (∀𝑥𝐴 ⊥ ↔ ¬ ∃𝑥𝐴 ⊤)
5 rextru 3069 . . 3 (∃𝑥 𝑥𝐴 ↔ ∃𝑥𝐴 ⊤)
65notbii 320 . 2 (¬ ∃𝑥 𝑥𝐴 ↔ ¬ ∃𝑥𝐴 ⊤)
7 ralfal.1 . . . 4 𝑥𝐴
87neq0f 4334 . . 3 𝐴 = ∅ ↔ ∃𝑥 𝑥𝐴)
98con1bii 356 . 2 (¬ ∃𝑥 𝑥𝐴𝐴 = ∅)
104, 6, 93bitr2ri 300 1 (𝐴 = ∅ ↔ ∀𝑥𝐴 ⊥)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 205   = wceq 1533  wtru 1534  wfal 1545  wex 1773  wcel 2098  wnfc 2875  wral 3053  wrex 3062  c0 4315
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-11 2146  ax-12 2163  ax-ext 2695
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1536  df-fal 1546  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2702  df-cleq 2716  df-clel 2802  df-nfc 2877  df-ral 3054  df-rex 3063  df-dif 3944  df-nul 4316
This theorem is referenced by: (None)
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