Theorem ssprr 3501

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 3232	. . . 4 ⊢ ∅ ⊆ {B, 𝐶}
2		sseq1 2943	. . . 4 ⊢ (A = ∅ → (A ⊆ {B, 𝐶} ↔ ∅ ⊆ {B, 𝐶}))
3	1, 2	mpbiri 157	. . 3 ⊢ (A = ∅ → A ⊆ {B, 𝐶})
4		snsspr1 3486	. . . 4 ⊢ {B} ⊆ {B, 𝐶}
5		sseq1 2943	. . . 4 ⊢ (A = {B} → (A ⊆ {B, 𝐶} ↔ {B} ⊆ {B, 𝐶}))
6	4, 5	mpbiri 157	. . 3 ⊢ (A = {B} → A ⊆ {B, 𝐶})
7	3, 6	jaoi 623	. 2 ⊢ ((A = ∅ ∨ A = {B}) → A ⊆ {B, 𝐶})
8		snsspr2 3487	. . . 4 ⊢ {𝐶} ⊆ {B, 𝐶}
9		sseq1 2943	. . . 4 ⊢ (A = {𝐶} → (A ⊆ {B, 𝐶} ↔ {𝐶} ⊆ {B, 𝐶}))
10	8, 9	mpbiri 157	. . 3 ⊢ (A = {𝐶} → A ⊆ {B, 𝐶})
11		eqimss 2974	. . 3 ⊢ (A = {B, 𝐶} → A ⊆ {B, 𝐶})
12	10, 11	jaoi 623	. 2 ⊢ ((A = {𝐶} ∨ A = {B, 𝐶}) → A ⊆ {B, 𝐶})
13	7, 12	jaoi 623	1 ⊢ (((A = ∅ ∨ A = {B}) ∨ (A = {𝐶} ∨ A = {B, 𝐶})) → A ⊆ {B, 𝐶})

Syntax hints:   → wi 4   ∨ wo 616   = wceq 1228   ⊆ wss 2894  ∅c0 3201  {csn 3350  {cpr 3351

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 532  ax-in2 533  ax-io 617  ax-5 1316  ax-7 1317  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-10 1377  ax-11 1378  ax-i12 1379  ax-bnd 1380  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-i5r 1410  ax-ext 2004

This theorem depends on definitions:  df-bi 110  df-tru 1231  df-nf 1330  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018  df-nfc 2149  df-v 2537  df-dif 2897  df-un 2899  df-in 2901  df-ss 2908  df-nul 3202  df-pr 3357

This theorem is referenced by:  sstpr  3502  pwprss  3550