Theorem sspr 4795

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 4349	. . . 4 ⊢ (∅ ∪ {𝐵, 𝐶}) = {𝐵, 𝐶}
2	1	sseq2i 3942	. . 3 ⊢ (𝐴 ⊆ (∅ ∪ {𝐵, 𝐶}) ↔ 𝐴 ⊆ {𝐵, 𝐶})
3		0ss 4353	. . . 4 ⊢ ∅ ⊆ 𝐴
4	3	biantrur 540	. . 3 ⊢ (𝐴 ⊆ (∅ ∪ {𝐵, 𝐶}) ↔ (∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})))
5	2, 4	bitr3i 280	. 2 ⊢ (𝐴 ⊆ {𝐵, 𝐶} ↔ (∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})))
6		ssunpr 4794	. 2 ⊢ ((∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})) ↔ ((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ∨ (𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶}))))
7		0un 4349	. . . . 5 ⊢ (∅ ∪ {𝐵}) = {𝐵}
8	7	eqeq2i 2776	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐵}) ↔ 𝐴 = {𝐵})
9	8	orbi2i 926	. . 3 ⊢ ((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ↔ (𝐴 = ∅ ∨ 𝐴 = {𝐵}))
10		0un 4349	. . . . 5 ⊢ (∅ ∪ {𝐶}) = {𝐶}
11	10	eqeq2i 2776	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐶}) ↔ 𝐴 = {𝐶})
12	1	eqeq2i 2776	. . . 4 ⊢ (𝐴 = (∅ ∪ {𝐵, 𝐶}) ↔ 𝐴 = {𝐵, 𝐶})
13	11, 12	orbi12i 928	. . 3 ⊢ ((𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶})) ↔ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶}))
14	9, 13	orbi12i 928	. 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})))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   ∨ wo 861   = wceq 1570   ⊆ wss 3900   ∪ cun 4152  ∅c0 4281  {csn 4588  {cpr 4590

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-sn 4589  df-pr 4591

This theorem is used by:  sstp  4796  pwpr  4833  propssopi  5487  indistopon  23374  bj-prmoore  38306