Theorem ssprr 3548

Description: The subsets of a pair. (Contributed by Jim Kingdon, 11-Aug-2018.)

Assertion

Ref	Expression
ssprr	⊢ (((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})) → 𝐴 ⊆ {𝐵, 𝐶})

Proof of Theorem ssprr

Step	Hyp	Ref	Expression
1		0ss 3282	. . . 4 ⊢ ∅ ⊆ {𝐵, 𝐶}
2		sseq1 3020	. . . 4 ⊢ (𝐴 = ∅ → (𝐴 ⊆ {𝐵, 𝐶} ↔ ∅ ⊆ {𝐵, 𝐶}))
3	1, 2	mpbiri 166	. . 3 ⊢ (𝐴 = ∅ → 𝐴 ⊆ {𝐵, 𝐶})
4		snsspr1 3533	. . . 4 ⊢ {𝐵} ⊆ {𝐵, 𝐶}
5		sseq1 3020	. . . 4 ⊢ (𝐴 = {𝐵} → (𝐴 ⊆ {𝐵, 𝐶} ↔ {𝐵} ⊆ {𝐵, 𝐶}))
6	4, 5	mpbiri 166	. . 3 ⊢ (𝐴 = {𝐵} → 𝐴 ⊆ {𝐵, 𝐶})
7	3, 6	jaoi 668	. 2 ⊢ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) → 𝐴 ⊆ {𝐵, 𝐶})
8		snsspr2 3534	. . . 4 ⊢ {𝐶} ⊆ {𝐵, 𝐶}
9		sseq1 3020	. . . 4 ⊢ (𝐴 = {𝐶} → (𝐴 ⊆ {𝐵, 𝐶} ↔ {𝐶} ⊆ {𝐵, 𝐶}))
10	8, 9	mpbiri 166	. . 3 ⊢ (𝐴 = {𝐶} → 𝐴 ⊆ {𝐵, 𝐶})
11		eqimss 3051	. . 3 ⊢ (𝐴 = {𝐵, 𝐶} → 𝐴 ⊆ {𝐵, 𝐶})
12	10, 11	jaoi 668	. 2 ⊢ ((𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶}) → 𝐴 ⊆ {𝐵, 𝐶})
13	7, 12	jaoi 668	1 ⊢ (((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})) → 𝐴 ⊆ {𝐵, 𝐶})

Syntax hints:   → wi 4   ∨ wo 661   = wceq 1284   ⊆ wss 2973  ∅c0 3251  {csn 3398  {cpr 3399

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 576  ax-in2 577  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603  df-dif 2975  df-un 2977  df-in 2979  df-ss 2986  df-nul 3252  df-pr 3405

This theorem is referenced by:  sstpr  3549  pwprss  3597