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Theorem undisjrab 40793
Description: Union of two disjoint restricted class abstractions; compare unrab 4252. (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 4314 . . 3 ({𝑥𝐴 ∣ (𝜑𝜓)} = ∅ ↔ ∀𝑥𝐴 ¬ (𝜑𝜓))
2 df-nan 1482 . . . . 5 ((𝜑𝜓) ↔ ¬ (𝜑𝜓))
3 nanorxor 40792 . . . . 5 ((𝜑𝜓) ↔ ((𝜑𝜓) ↔ (𝜑𝜓)))
42, 3bitr3i 279 . . . 4 (¬ (𝜑𝜓) ↔ ((𝜑𝜓) ↔ (𝜑𝜓)))
54ralbii 3152 . . 3 (∀𝑥𝐴 ¬ (𝜑𝜓) ↔ ∀𝑥𝐴 ((𝜑𝜓) ↔ (𝜑𝜓)))
6 rabbi 3370 . . 3 (∀𝑥𝐴 ((𝜑𝜓) ↔ (𝜑𝜓)) ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
71, 5, 63bitri 299 . 2 ({𝑥𝐴 ∣ (𝜑𝜓)} = ∅ ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
8 inrab 4253 . . 3 ({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)}
98eqeq1i 2825 . 2 (({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = ∅ ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = ∅)
10 unrab 4252 . . 3 ({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)}
1110eqeq1i 2825 . 2 (({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)} ↔ {𝑥𝐴 ∣ (𝜑𝜓)} = {𝑥𝐴 ∣ (𝜑𝜓)})
127, 9, 113bitr4i 305 1 (({𝑥𝐴𝜑} ∩ {𝑥𝐴𝜓}) = ∅ ↔ ({𝑥𝐴𝜑} ∪ {𝑥𝐴𝜓}) = {𝑥𝐴 ∣ (𝜑𝜓)})
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 208  wa 398  wo 843  wnan 1481  wxo 1501   = wceq 1537  wral 3125  {crab 3129  cun 3911  cin 3912  c0 4269
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2792
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-nan 1482  df-xor 1502  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2799  df-cleq 2813  df-clel 2891  df-nfc 2959  df-ral 3130  df-rab 3134  df-v 3475  df-dif 3916  df-un 3918  df-in 3920  df-nul 4270
This theorem is referenced by: (None)
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