Theorem ssprr 3527

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 3255	. . . 4 ⊢ ∅ ⊆ {𝐵, 𝐶}
2		sseq1 2966	. . . 4 ⊢ (𝐴 = ∅ → (𝐴 ⊆ {𝐵, 𝐶} ↔ ∅ ⊆ {𝐵, 𝐶}))
3	1, 2	mpbiri 157	. . 3 ⊢ (𝐴 = ∅ → 𝐴 ⊆ {𝐵, 𝐶})
4		snsspr1 3512	. . . 4 ⊢ {𝐵} ⊆ {𝐵, 𝐶}
5		sseq1 2966	. . . 4 ⊢ (𝐴 = {𝐵} → (𝐴 ⊆ {𝐵, 𝐶} ↔ {𝐵} ⊆ {𝐵, 𝐶}))
6	4, 5	mpbiri 157	. . 3 ⊢ (𝐴 = {𝐵} → 𝐴 ⊆ {𝐵, 𝐶})
7	3, 6	jaoi 636	. 2 ⊢ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) → 𝐴 ⊆ {𝐵, 𝐶})
8		snsspr2 3513	. . . 4 ⊢ {𝐶} ⊆ {𝐵, 𝐶}
9		sseq1 2966	. . . 4 ⊢ (𝐴 = {𝐶} → (𝐴 ⊆ {𝐵, 𝐶} ↔ {𝐶} ⊆ {𝐵, 𝐶}))
10	8, 9	mpbiri 157	. . 3 ⊢ (𝐴 = {𝐶} → 𝐴 ⊆ {𝐵, 𝐶})
11		eqimss 2997	. . 3 ⊢ (𝐴 = {𝐵, 𝐶} → 𝐴 ⊆ {𝐵, 𝐶})
12	10, 11	jaoi 636	. 2 ⊢ ((𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶}) → 𝐴 ⊆ {𝐵, 𝐶})
13	7, 12	jaoi 636	1 ⊢ (((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})) → 𝐴 ⊆ {𝐵, 𝐶})

Syntax hints:   → wi 4   ∨ wo 629   = wceq 1243   ⊆ wss 2917  ∅c0 3224  {csn 3375  {cpr 3376

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559  df-dif 2920  df-un 2922  df-in 2924  df-ss 2931  df-nul 3225  df-pr 3382

This theorem is referenced by:  sstpr  3528  pwprss  3576