Theorem undisjrab 44875

Description: Union of two disjoint restricted class abstractions; compare unrab 4270. (Contributed by Steve Rodriguez, 28-Feb-2020.)

Assertion

Ref	Expression
undisjrab	⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ∅ ↔ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)})

Proof of Theorem undisjrab

Step	Hyp	Ref	Expression
1		rabeq0 4345	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = ∅ ↔ ∀𝑥 ∈ 𝐴 ¬ (𝜑 ∧ 𝜓))
2		df-nan 1515	. . . . 5 ⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓))
3		nanorxor 44874	. . . . 5 ⊢ ((𝜑 ⊼ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
4	2, 3	bitr3i 280	. . . 4 ⊢ (¬ (𝜑 ∧ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
5	4	ralbii 3111	. . 3 ⊢ (∀𝑥 ∈ 𝐴 ¬ (𝜑 ∧ 𝜓) ↔ ∀𝑥 ∈ 𝐴 ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
6		rabbi 3447	. . 3 ⊢ (∀𝑥 ∈ 𝐴 ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)) ↔ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)})
7	1, 5, 6	3bitri 300	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = ∅ ↔ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)})
8		inrab 4271	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}
9	8	eqeq1i 2770	. 2 ⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ∅ ↔ {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = ∅)
10		unrab 4270	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)}
11	10	eqeq1i 2770	. 2 ⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)} ↔ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)})
12	7, 9, 11	3bitr4i 306	1 ⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ∅ ↔ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)})

Syntax hints:  ¬ wn 3   ↔ wb 209   ∧ wa 400   ∨ wo 860   ⊼ wnan 1514   ⊻ wxo 1534   = wceq 1563  ∀wral 3079  {crab 3417   ∪ cun 3905   ∩ cin 3906  ∅c0 4288

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2737

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-nan 1515  df-xor 1535  df-tru 1566  df-fal 1576  df-ex 1803  df-nf 1807  df-sb 2094  df-clab 2744  df-cleq 2757  df-clel 2840  df-ral 3080  df-rab 3418  df-v 3459  df-dif 3910  df-un 3912  df-in 3914  df-nul 4289

This theorem is referenced by: (None)