Theorem undisjrab 44275

Description: Union of two disjoint restricted class abstractions; compare unrab 4334. (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 4411	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = ∅ ↔ ∀𝑥 ∈ 𝐴 ¬ (𝜑 ∧ 𝜓))
2		df-nan 1489	. . . . 5 ⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓))
3		nanorxor 44274	. . . . 5 ⊢ ((𝜑 ⊼ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
4	2, 3	bitr3i 277	. . . 4 ⊢ (¬ (𝜑 ∧ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
5	4	ralbii 3099	. . 3 ⊢ (∀𝑥 ∈ 𝐴 ¬ (𝜑 ∧ 𝜓) ↔ ∀𝑥 ∈ 𝐴 ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
6		rabbi 3475	. . 3 ⊢ (∀𝑥 ∈ 𝐴 ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)) ↔ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)})
7	1, 5, 6	3bitri 297	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = ∅ ↔ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)})
8		inrab 4335	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}
9	8	eqeq1i 2745	. 2 ⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ∅ ↔ {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = ∅)
10		unrab 4334	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)}
11	10	eqeq1i 2745	. 2 ⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)} ↔ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)})
12	7, 9, 11	3bitr4i 303	1 ⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ∅ ↔ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)})

Syntax hints:  ¬ wn 3   ↔ wb 206   ∧ wa 395   ∨ wo 846   ⊼ wnan 1488   ⊻ wxo 1508   = wceq 1537  ∀wral 3067  {crab 3443   ∪ cun 3974   ∩ cin 3975  ∅c0 4352

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2158  ax-12 2178  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-nan 1489  df-xor 1509  df-tru 1540  df-fal 1550  df-ex 1778  df-nf 1782  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ral 3068  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-in 3983  df-nul 4353

This theorem is referenced by: (None)