Theorem sspr 4779

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 4337	. . . 4 ⊢ (∅ ∪ {𝐵, 𝐶}) = {𝐵, 𝐶}
2	1	sseq2i 3952	. . 3 ⊢ (𝐴 ⊆ (∅ ∪ {𝐵, 𝐶}) ↔ 𝐴 ⊆ {𝐵, 𝐶})
3		0ss 4341	. . . 4 ⊢ ∅ ⊆ 𝐴
4	3	biantrur 530	. . 3 ⊢ (𝐴 ⊆ (∅ ∪ {𝐵, 𝐶}) ↔ (∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})))
5	2, 4	bitr3i 277	. 2 ⊢ (𝐴 ⊆ {𝐵, 𝐶} ↔ (∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})))
6		ssunpr 4778	. 2 ⊢ ((∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})) ↔ ((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ∨ (𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶}))))
7		0un 4337	. . . . 5 ⊢ (∅ ∪ {𝐵}) = {𝐵}
8	7	eqeq2i 2750	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐵}) ↔ 𝐴 = {𝐵})
9	8	orbi2i 913	. . 3 ⊢ ((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ↔ (𝐴 = ∅ ∨ 𝐴 = {𝐵}))
10		0un 4337	. . . . 5 ⊢ (∅ ∪ {𝐶}) = {𝐶}
11	10	eqeq2i 2750	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐶}) ↔ 𝐴 = {𝐶})
12	1	eqeq2i 2750	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐵, 𝐶}) ↔ 𝐴 = {𝐵, 𝐶})
13	11, 12	orbi12i 915	. . 3 ⊢ ((𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶})) ↔ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶}))
14	9, 13	orbi12i 915	. 2 ⊢ (((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ∨ (𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶}))) ↔ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})))
15	5, 6, 14	3bitri 297	1 ⊢ (𝐴 ⊆ {𝐵, 𝐶} ↔ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})))

Syntax hints:   ↔ wb 206   ∧ wa 395   ∨ wo 848   = wceq 1542   ∪ cun 3888   ⊆ wss 3890  ∅c0 4274  {csn 4568  {cpr 4570

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ral 3053  df-v 3432  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-nul 4275  df-sn 4569  df-pr 4571

This theorem is referenced by:  sstp  4780  pwpr  4845  propssopi  5457  indistopon  22979  bj-prmoore  37446