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Theorem ralfal 45994
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 1583 . . . 4 (⊥ ↔ ¬ ⊤)
21ralbii 3108 . . 3 (∀𝑥𝐴 ⊥ ↔ ∀𝑥𝐴 ¬ ⊤)
3 ralnex 3088 . . 3 (∀𝑥𝐴 ¬ ⊤ ↔ ¬ ∃𝑥𝐴 ⊤)
42, 3bitri 278 . 2 (∀𝑥𝐴 ⊥ ↔ ¬ ∃𝑥𝐴 ⊤)
5 rextru 3093 . . 3 (∃𝑥 𝑥𝐴 ↔ ∃𝑥𝐴 ⊤)
65notbii 323 . 2 (¬ ∃𝑥 𝑥𝐴 ↔ ¬ ∃𝑥𝐴 ⊤)
7 ralfal.1 . . . 4 𝑥𝐴
87neq0f 4295 . . 3 𝐴 = ∅ ↔ ∃𝑥 𝑥𝐴)
98con1bii 359 . 2 (¬ ∃𝑥 𝑥𝐴𝐴 = ∅)
104, 6, 93bitr2ri 303 1 (𝐴 = ∅ ↔ ∀𝑥𝐴 ⊥)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3  wb 209   = wceq 1570  wtru 1571  wfal 1582  wex 1812  wcel 2145  wnfc 2907  wral 3076  wrex 3086  c0 4279
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-11 2194  ax-12 2213  ax-ext 2732
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ral 3077  df-rex 3087  df-dif 3902  df-nul 4280
This theorem is used by: (None)
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