Theorem undisjrab 40631

Description: Union of two disjoint restricted class abstractions; compare unrab 4273. (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 4337	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = ∅ ↔ ∀𝑥 ∈ 𝐴 ¬ (𝜑 ∧ 𝜓))
2		df-nan 1481	. . . . 5 ⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓))
3		nanorxor 40630	. . . . 5 ⊢ ((𝜑 ⊼ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
4	2, 3	bitr3i 279	. . . 4 ⊢ (¬ (𝜑 ∧ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
5	4	ralbii 3165	. . 3 ⊢ (∀𝑥 ∈ 𝐴 ¬ (𝜑 ∧ 𝜓) ↔ ∀𝑥 ∈ 𝐴 ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
6		rabbi 3383	. . 3 ⊢ (∀𝑥 ∈ 𝐴 ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)) ↔ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)})
7	1, 5, 6	3bitri 299	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = ∅ ↔ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)})
8		inrab 4274	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}
9	8	eqeq1i 2826	. 2 ⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ∅ ↔ {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = ∅)
10		unrab 4273	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)}
11	10	eqeq1i 2826	. 2 ⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)} ↔ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)})
12	7, 9, 11	3bitr4i 305	1 ⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ∅ ↔ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ⊻ 𝜓)})

Syntax hints:  ¬ wn 3   ↔ wb 208   ∧ wa 398   ∨ wo 843   ⊼ wnan 1480   ⊻ wxo 1500   = wceq 1533  ∀wral 3138  {crab 3142   ∪ cun 3933   ∩ cin 3934  ∅c0 4290

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-nan 1481  df-xor 1501  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-ral 3143  df-rab 3147  df-v 3496  df-dif 3938  df-un 3940  df-in 3942  df-nul 4291

This theorem is referenced by: (None)