Theorem undisjrab 45384

Description: Union of two disjoint restricted class abstractions; compare unrab 4260. (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 1521	. . . . 5 ⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓))
3		nanorxor 45383	. . . . 5 ⊢ ((𝜑 ⊼ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
4	2, 3	bitr3i 280	. . . 4 ⊢ (¬ (𝜑 ∧ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
5	4	ralbii 3108	. . 3 ⊢ (∀𝑥 ∈ 𝐴 ¬ (𝜑 ∧ 𝜓) ↔ ∀𝑥 ∈ 𝐴 ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
6		rabbi 3441	. . 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 4261	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}
9	8	eqeq1i 2765	. 2 ⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ∅ ↔ {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = ∅)
10		unrab 4260	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)}
11	10	eqeq1i 2765	. 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)})

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   ∧ wa 401   ∨ wo 861   ⊼ wnan 1520   ⊻ wxo 1540   = wceq 1569  ∀wral 3076  {crab 3412   ∩ cin 4089   ∪ cun 4149  ∅c0 4278

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-nan 1521  df-xor 1541  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rab 3413  df-v 3452  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279

This theorem is used by: (None)