Theorem sspr 4804

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 4360	. . . 4 ⊢ (∅ ∪ {𝐵, 𝐶}) = {𝐵, 𝐶}
2	1	sseq2i 3974	. . 3 ⊢ (𝐴 ⊆ (∅ ∪ {𝐵, 𝐶}) ↔ 𝐴 ⊆ {𝐵, 𝐶})
3		0ss 4364	. . . 4 ⊢ ∅ ⊆ 𝐴
4	3	biantrur 539	. . 3 ⊢ (𝐴 ⊆ (∅ ∪ {𝐵, 𝐶}) ↔ (∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})))
5	2, 4	bitr3i 280	. 2 ⊢ (𝐴 ⊆ {𝐵, 𝐶} ↔ (∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})))
6		ssunpr 4803	. 2 ⊢ ((∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})) ↔ ((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ∨ (𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶}))))
7		0un 4360	. . . . 5 ⊢ (∅ ∪ {𝐵}) = {𝐵}
8	7	eqeq2i 2782	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐵}) ↔ 𝐴 = {𝐵})
9	8	orbi2i 925	. . 3 ⊢ ((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ↔ (𝐴 = ∅ ∨ 𝐴 = {𝐵}))
10		0un 4360	. . . . 5 ⊢ (∅ ∪ {𝐶}) = {𝐶}
11	10	eqeq2i 2782	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐶}) ↔ 𝐴 = {𝐶})
12	1	eqeq2i 2782	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐵, 𝐶}) ↔ 𝐴 = {𝐵, 𝐶})
13	11, 12	orbi12i 927	. . 3 ⊢ ((𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶})) ↔ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶}))
14	9, 13	orbi12i 927	. 2 ⊢ (((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ∨ (𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶}))) ↔ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})))
15	5, 6, 14	3bitri 300	1 ⊢ (𝐴 ⊆ {𝐵, 𝐶} ↔ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})))

Syntax hints:   ↔ wb 209   ∧ wa 400   ∨ wo 860   = wceq 1567   ∪ cun 3911   ⊆ wss 3913  ∅c0 4294  {csn 4594  {cpr 4596

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ral 3086  df-v 3465  df-dif 3916  df-un 3918  df-in 3920  df-ss 3930  df-nul 4295  df-sn 4595  df-pr 4597

This theorem is referenced by:  sstp  4805  pwpr  4870  propssopi  5492  indistopon  23126  bj-prmoore  37644