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Theorem undisjrab 38822
Description: Union of two disjoint restricted class abstractions; compare unrab 3931. (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 3990 . . 3 ({𝑥𝐴 ∣ (𝜑𝜓)} = ∅ ↔ ∀𝑥𝐴 ¬ (𝜑𝜓))
2 df-nan 1488 . . . . 5 ((𝜑𝜓) ↔ ¬ (𝜑𝜓))
3 nanorxor 38821 . . . . 5 ((𝜑𝜓) ↔ ((𝜑𝜓) ↔ (𝜑𝜓)))
42, 3bitr3i 266 . . . 4 (¬ (𝜑𝜓) ↔ ((𝜑𝜓) ↔ (𝜑𝜓)))
54ralbii 3009 . . 3 (∀𝑥𝐴 ¬ (𝜑𝜓) ↔ ∀𝑥𝐴 ((𝜑𝜓) ↔ (𝜑𝜓)))
6 rabbi 3150 . . 3 (∀𝑥𝐴 ((𝜑𝜓) ↔ (𝜑𝜓)) ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
71, 5, 63bitri 286 . 2 ({𝑥𝐴 ∣ (𝜑𝜓)} = ∅ ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
8 inrab 3932 . . 3 ({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)}
98eqeq1i 2656 . 2 (({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = ∅ ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = ∅)
10 unrab 3931 . . 3 ({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)}
1110eqeq1i 2656 . 2 (({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)} ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
127, 9, 113bitr4i 292 1 (({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = ∅ ↔ ({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)})
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 196  wo 382  wa 383  wnan 1487  wxo 1504   = wceq 1523  wral 2941  {crab 2945  cun 3605  cin 3606  c0 3948
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-13 2282  ax-ext 2631
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-nan 1488  df-xor 1505  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-clab 2638  df-cleq 2644  df-clel 2647  df-nfc 2782  df-ral 2946  df-rab 2950  df-v 3233  df-dif 3610  df-un 3612  df-in 3614  df-nul 3949
This theorem is referenced by: (None)
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