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Theorem undisjrab 43622
Description: Union of two disjoint restricted class abstractions; compare unrab 4300. (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 4379 . . 3 ({𝑥𝐴 ∣ (𝜑𝜓)} = ∅ ↔ ∀𝑥𝐴 ¬ (𝜑𝜓))
2 df-nan 1485 . . . . 5 ((𝜑𝜓) ↔ ¬ (𝜑𝜓))
3 nanorxor 43621 . . . . 5 ((𝜑𝜓) ↔ ((𝜑𝜓) ↔ (𝜑𝜓)))
42, 3bitr3i 277 . . . 4 (¬ (𝜑𝜓) ↔ ((𝜑𝜓) ↔ (𝜑𝜓)))
54ralbii 3087 . . 3 (∀𝑥𝐴 ¬ (𝜑𝜓) ↔ ∀𝑥𝐴 ((𝜑𝜓) ↔ (𝜑𝜓)))
6 rabbi 3456 . . 3 (∀𝑥𝐴 ((𝜑𝜓) ↔ (𝜑𝜓)) ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
71, 5, 63bitri 297 . 2 ({𝑥𝐴 ∣ (𝜑𝜓)} = ∅ ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
8 inrab 4301 . . 3 ({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)}
98eqeq1i 2731 . 2 (({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = ∅ ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = ∅)
10 unrab 4300 . . 3 ({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)}
1110eqeq1i 2731 . 2 (({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)} ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
127, 9, 113bitr4i 303 1 (({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = ∅ ↔ ({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)})
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 205  wa 395  wo 844  wnan 1484  wxo 1504   = wceq 1533  wral 3055  {crab 3426  cun 3941  cin 3942  c0 4317
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-10 2129  ax-11 2146  ax-12 2163  ax-ext 2697
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-nan 1485  df-xor 1505  df-tru 1536  df-fal 1546  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2704  df-cleq 2718  df-clel 2804  df-ral 3056  df-rab 3427  df-v 3470  df-dif 3946  df-un 3948  df-in 3950  df-nul 4318
This theorem is referenced by: (None)
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