Theorem sspr 4793

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 4350	. . . 4 ⊢ (∅ ∪ {𝐵, 𝐶}) = {𝐵, 𝐶}
2	1	sseq2i 3965	. . 3 ⊢ (𝐴 ⊆ (∅ ∪ {𝐵, 𝐶}) ↔ 𝐴 ⊆ {𝐵, 𝐶})
3		0ss 4354	. . . 4 ⊢ ∅ ⊆ 𝐴
4	3	biantrur 538	. . 3 ⊢ (𝐴 ⊆ (∅ ∪ {𝐵, 𝐶}) ↔ (∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})))
5	2, 4	bitr3i 279	. 2 ⊢ (𝐴 ⊆ {𝐵, 𝐶} ↔ (∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})))
6		ssunpr 4792	. 2 ⊢ ((∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})) ↔ ((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ∨ (𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶}))))
7		0un 4350	. . . . 5 ⊢ (∅ ∪ {𝐵}) = {𝐵}
8	7	eqeq2i 2775	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐵}) ↔ 𝐴 = {𝐵})
9	8	orbi2i 923	. . 3 ⊢ ((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ↔ (𝐴 = ∅ ∨ 𝐴 = {𝐵}))
10		0un 4350	. . . . 5 ⊢ (∅ ∪ {𝐶}) = {𝐶}
11	10	eqeq2i 2775	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐶}) ↔ 𝐴 = {𝐶})
12	1	eqeq2i 2775	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐵, 𝐶}) ↔ 𝐴 = {𝐵, 𝐶})
13	11, 12	orbi12i 925	. . 3 ⊢ ((𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶})) ↔ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶}))
14	9, 13	orbi12i 925	. 2 ⊢ (((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ∨ (𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶}))) ↔ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})))
15	5, 6, 14	3bitri 299	1 ⊢ (𝐴 ⊆ {𝐵, 𝐶} ↔ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})))

Syntax hints:   ↔ wb 208   ∧ wa 399   ∨ wo 858   = wceq 1560   ∪ cun 3902   ⊆ wss 3904  ∅c0 4285  {csn 4582  {cpr 4584

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-ext 2734

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1563  df-fal 1573  df-ex 1800  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-ral 3077  df-v 3456  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-sn 4583  df-pr 4585

This theorem is referenced by:  sstp  4794  pwpr  4859  propssopi  5477  indistopon  23061  bj-prmoore  37605