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Theorem undisjrab 37321
Description: Union of two disjoint restricted class abstractions; compare unrab 3857. (Contributed by Steve Rodriguez, 28-Feb-2020.)
Assertion
Ref Expression
undisjrab (({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = ∅ ↔ ({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)})

Proof of Theorem undisjrab
StepHypRef Expression
1 rabeq0 3911 . . 3 ({𝑥𝐴 ∣ (𝜑𝜓)} = ∅ ↔ ∀𝑥𝐴 ¬ (𝜑𝜓))
2 df-nan 1440 . . . . 5 ((𝜑𝜓) ↔ ¬ (𝜑𝜓))
3 nanorxor 37320 . . . . 5 ((𝜑𝜓) ↔ ((𝜑𝜓) ↔ (𝜑𝜓)))
42, 3bitr3i 265 . . . 4 (¬ (𝜑𝜓) ↔ ((𝜑𝜓) ↔ (𝜑𝜓)))
54ralbii 2963 . . 3 (∀𝑥𝐴 ¬ (𝜑𝜓) ↔ ∀𝑥𝐴 ((𝜑𝜓) ↔ (𝜑𝜓)))
6 rabbi 3097 . . 3 (∀𝑥𝐴 ((𝜑𝜓) ↔ (𝜑𝜓)) ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
71, 5, 63bitri 285 . 2 ({𝑥𝐴 ∣ (𝜑𝜓)} = ∅ ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
8 inrab 3858 . . 3 ({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)}
98eqeq1i 2615 . 2 (({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = ∅ ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = ∅)
10 unrab 3857 . . 3 ({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)}
1110eqeq1i 2615 . 2 (({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)} ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
127, 9, 113bitr4i 291 1 (({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = ∅ ↔ ({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)})
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 195  wo 382  wa 383  wnan 1439  wxo 1456   = wceq 1475  wral 2896  {crab 2900  cun 3538  cin 3539  c0 3874
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590
This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-nan 1440  df-xor 1457  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ral 2901  df-rab 2905  df-v 3175  df-dif 3543  df-un 3545  df-in 3547  df-nul 3875
This theorem is referenced by: (None)
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