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Theorem ralfal 45066
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 1550 . . . 4 (⊥ ↔ ¬ ⊤)
21ralbii 3099 . . 3 (∀𝑥𝐴 ⊥ ↔ ∀𝑥𝐴 ¬ ⊤)
3 ralnex 3078 . . 3 (∀𝑥𝐴 ¬ ⊤ ↔ ¬ ∃𝑥𝐴 ⊤)
42, 3bitri 275 . 2 (∀𝑥𝐴 ⊥ ↔ ¬ ∃𝑥𝐴 ⊤)
5 rextru 3083 . . 3 (∃𝑥 𝑥𝐴 ↔ ∃𝑥𝐴 ⊤)
65notbii 320 . 2 (¬ ∃𝑥 𝑥𝐴 ↔ ¬ ∃𝑥𝐴 ⊤)
7 ralfal.1 . . . 4 𝑥𝐴
87neq0f 4371 . . 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 206   = wceq 1537  wtru 1538  wfal 1549  wex 1777  wcel 2108  wnfc 2893  wral 3067  wrex 3076  c0 4352
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-11 2158  ax-12 2178  ax-ext 2711
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-tru 1540  df-fal 1550  df-ex 1778  df-nf 1782  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-nfc 2895  df-ral 3068  df-rex 3077  df-dif 3979  df-nul 4353
This theorem is referenced by: (None)
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