Theorem sspr 4766

Description: The subsets of a pair. (Contributed by NM, 16-Mar-2006.) (Proof shortened by Mario Carneiro, 2-Jul-2016.)

Assertion

Ref	Expression
sspr	⊢ (𝐴 ⊆ {𝐵, 𝐶} ↔ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})))

Proof of Theorem sspr

Step	Hyp	Ref	Expression
1		0un 4324	. . . 4 ⊢ (∅ ∪ {𝐵, 𝐶}) = {𝐵, 𝐶}
2	1	sseq2i 3944	. . 3 ⊢ (𝐴 ⊆ (∅ ∪ {𝐵, 𝐶}) ↔ 𝐴 ⊆ {𝐵, 𝐶})
3		0ss 4328	. . . 4 ⊢ ∅ ⊆ 𝐴
4	3	biantrur 535	. . 3 ⊢ (𝐴 ⊆ (∅ ∪ {𝐵, 𝐶}) ↔ (∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})))
5	2, 4	bitr3i 278	. 2 ⊢ (𝐴 ⊆ {𝐵, 𝐶} ↔ (∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})))
6		ssunpr 4765	. 2 ⊢ ((∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})) ↔ ((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ∨ (𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶}))))
7		0un 4324	. . . . 5 ⊢ (∅ ∪ {𝐵}) = {𝐵}
8	7	eqeq2i 2752	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐵}) ↔ 𝐴 = {𝐵})
9	8	orbi2i 918	. . 3 ⊢ ((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ↔ (𝐴 = ∅ ∨ 𝐴 = {𝐵}))
10		0un 4324	. . . . 5 ⊢ (∅ ∪ {𝐶}) = {𝐶}
11	10	eqeq2i 2752	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐶}) ↔ 𝐴 = {𝐶})
12	1	eqeq2i 2752	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐵, 𝐶}) ↔ 𝐴 = {𝐵, 𝐶})
13	11, 12	orbi12i 920	. . 3 ⊢ ((𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶})) ↔ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶}))
14	9, 13	orbi12i 920	. 2 ⊢ (((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ∨ (𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶}))) ↔ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})))
15	5, 6, 14	3bitri 298	1 ⊢ (𝐴 ⊆ {𝐵, 𝐶} ↔ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})))

Syntax hints:   ↔ wb 207   ∧ wa 396   ∨ wo 853   = wceq 1547   ∪ cun 3881   ⊆ wss 3883  ∅c0 4261  {csn 4555  {cpr 4557

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-ral 3054  df-v 3433  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4262  df-sn 4556  df-pr 4558

This theorem is referenced by:  sstp  4767  pwpr  4832  propssopi  5449  indistopon  22984  bj-prmoore  37473