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Theorem undisjrab 43746
Description: Union of two disjoint restricted class abstractions; compare unrab 4306. (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 4386 . . 3 ({𝑥𝐴 ∣ (𝜑𝜓)} = ∅ ↔ ∀𝑥𝐴 ¬ (𝜑𝜓))
2 df-nan 1485 . . . . 5 ((𝜑𝜓) ↔ ¬ (𝜑𝜓))
3 nanorxor 43745 . . . . 5 ((𝜑𝜓) ↔ ((𝜑𝜓) ↔ (𝜑𝜓)))
42, 3bitr3i 276 . . . 4 (¬ (𝜑𝜓) ↔ ((𝜑𝜓) ↔ (𝜑𝜓)))
54ralbii 3089 . . 3 (∀𝑥𝐴 ¬ (𝜑𝜓) ↔ ∀𝑥𝐴 ((𝜑𝜓) ↔ (𝜑𝜓)))
6 rabbi 3459 . . 3 (∀𝑥𝐴 ((𝜑𝜓) ↔ (𝜑𝜓)) ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
71, 5, 63bitri 296 . 2 ({𝑥𝐴 ∣ (𝜑𝜓)} = ∅ ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
8 inrab 4307 . . 3 ({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)}
98eqeq1i 2732 . 2 (({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = ∅ ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = ∅)
10 unrab 4306 . . 3 ({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)}
1110eqeq1i 2732 . 2 (({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)} ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
127, 9, 113bitr4i 302 1 (({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = ∅ ↔ ({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)})
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 205  wa 394  wo 845  wnan 1484  wxo 1504   = wceq 1533  wral 3057  {crab 3428  cun 3945  cin 3946  c0 4324
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 2166  ax-ext 2698
This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-nan 1485  df-xor 1505  df-tru 1536  df-fal 1546  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2705  df-cleq 2719  df-clel 2805  df-ral 3058  df-rab 3429  df-v 3473  df-dif 3950  df-un 3952  df-in 3954  df-nul 4325
This theorem is referenced by: (None)
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