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Theorem ralfal 45608
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 1560 . . . 4 (⊥ ↔ ¬ ⊤)
21ralbii 3085 . . 3 (∀𝑥𝐴 ⊥ ↔ ∀𝑥𝐴 ¬ ⊤)
3 ralnex 3065 . . 3 (∀𝑥𝐴 ¬ ⊤ ↔ ¬ ∃𝑥𝐴 ⊤)
42, 3bitri 276 . 2 (∀𝑥𝐴 ⊥ ↔ ¬ ∃𝑥𝐴 ⊤)
5 rextru 3070 . . 3 (∃𝑥 𝑥𝐴 ↔ ∃𝑥𝐴 ⊤)
65notbii 321 . 2 (¬ ∃𝑥 𝑥𝐴 ↔ ¬ ∃𝑥𝐴 ⊤)
7 ralfal.1 . . . 4 𝑥𝐴
87neq0f 4276 . . 3 𝐴 = ∅ ↔ ∃𝑥 𝑥𝐴)
98con1bii 357 . 2 (¬ ∃𝑥 𝑥𝐴𝐴 = ∅)
104, 6, 93bitr2ri 301 1 (𝐴 = ∅ ↔ ∀𝑥𝐴 ⊥)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 207   = wceq 1547  wtru 1548  wfal 1559  wex 1786  wcel 2119  wnfc 2886  wral 3053  wrex 3063  c0 4261
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-11 2168  ax-12 2189  ax-ext 2711
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-tru 1550  df-fal 1560  df-ex 1787  df-nf 1791  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-nfc 2888  df-ral 3054  df-rex 3064  df-dif 3886  df-nul 4262
This theorem is referenced by: (None)
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