Theorem ssprr 3518

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 3249	. . . 4 ⊢ ∅ ⊆ {B, 𝐶}
2		sseq1 2960	. . . 4 ⊢ (A = ∅ → (A ⊆ {B, 𝐶} ↔ ∅ ⊆ {B, 𝐶}))
3	1, 2	mpbiri 157	. . 3 ⊢ (A = ∅ → A ⊆ {B, 𝐶})
4		snsspr1 3503	. . . 4 ⊢ {B} ⊆ {B, 𝐶}
5		sseq1 2960	. . . 4 ⊢ (A = {B} → (A ⊆ {B, 𝐶} ↔ {B} ⊆ {B, 𝐶}))
6	4, 5	mpbiri 157	. . 3 ⊢ (A = {B} → A ⊆ {B, 𝐶})
7	3, 6	jaoi 635	. 2 ⊢ ((A = ∅ ∨ A = {B}) → A ⊆ {B, 𝐶})
8		snsspr2 3504	. . . 4 ⊢ {𝐶} ⊆ {B, 𝐶}
9		sseq1 2960	. . . 4 ⊢ (A = {𝐶} → (A ⊆ {B, 𝐶} ↔ {𝐶} ⊆ {B, 𝐶}))
10	8, 9	mpbiri 157	. . 3 ⊢ (A = {𝐶} → A ⊆ {B, 𝐶})
11		eqimss 2991	. . 3 ⊢ (A = {B, 𝐶} → A ⊆ {B, 𝐶})
12	10, 11	jaoi 635	. 2 ⊢ ((A = {𝐶} ∨ A = {B, 𝐶}) → A ⊆ {B, 𝐶})
13	7, 12	jaoi 635	1 ⊢ (((A = ∅ ∨ A = {B}) ∨ (A = {𝐶} ∨ A = {B, 𝐶})) → A ⊆ {B, 𝐶})

Syntax hints:   → wi 4   ∨ wo 628   = wceq 1242   ⊆ wss 2911  ∅c0 3218  {csn 3367  {cpr 3368

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 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-dif 2914  df-un 2916  df-in 2918  df-ss 2925  df-nul 3219  df-pr 3374

This theorem is referenced by:  sstpr  3519  pwprss  3567