Theorem undisjrab 44842

Description: Union of two disjoint restricted class abstractions; compare unrab 4265. (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 4339	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = ∅ ↔ ∀𝑥 ∈ 𝐴 ¬ (𝜑 ∧ 𝜓))
2		df-nan 1511	. . . . 5 ⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓))
3		nanorxor 44841	. . . . 5 ⊢ ((𝜑 ⊼ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
4	2, 3	bitr3i 279	. . . 4 ⊢ (¬ (𝜑 ∧ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
5	4	ralbii 3107	. . 3 ⊢ (∀𝑥 ∈ 𝐴 ¬ (𝜑 ∧ 𝜓) ↔ ∀𝑥 ∈ 𝐴 ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
6		rabbi 3443	. . 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 4266	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}
9	8	eqeq1i 2766	. 2 ⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ∅ ↔ {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = ∅)
10		unrab 4265	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)}
11	10	eqeq1i 2766	. 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 399   ∨ wo 858   ⊼ wnan 1510   ⊻ wxo 1530   = wceq 1559  ∀wral 3075  {crab 3413   ∪ cun 3900   ∩ cin 3901  ∅c0 4283

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-10 2174  ax-11 2190  ax-12 2211  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-nan 1511  df-xor 1531  df-tru 1562  df-fal 1572  df-ex 1799  df-nf 1803  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3076  df-rab 3414  df-v 3455  df-dif 3905  df-un 3907  df-in 3909  df-nul 4284

This theorem is referenced by: (None)