Theorem ssprr 3879

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 3561	. . . 4 ⊢ ∅ ⊆ {𝐵, 𝐶}
2		sseq1 3271	. . . 4 ⊢ (𝐴 = ∅ → (𝐴 ⊆ {𝐵, 𝐶} ↔ ∅ ⊆ {𝐵, 𝐶}))
3	1, 2	mpbiri 168	. . 3 ⊢ (𝐴 = ∅ → 𝐴 ⊆ {𝐵, 𝐶})
4		snsspr1 3861	. . . 4 ⊢ {𝐵} ⊆ {𝐵, 𝐶}
5		sseq1 3271	. . . 4 ⊢ (𝐴 = {𝐵} → (𝐴 ⊆ {𝐵, 𝐶} ↔ {𝐵} ⊆ {𝐵, 𝐶}))
6	4, 5	mpbiri 168	. . 3 ⊢ (𝐴 = {𝐵} → 𝐴 ⊆ {𝐵, 𝐶})
7	3, 6	jaoi 728	. 2 ⊢ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) → 𝐴 ⊆ {𝐵, 𝐶})
8		snsspr2 3862	. . . 4 ⊢ {𝐶} ⊆ {𝐵, 𝐶}
9		sseq1 3271	. . . 4 ⊢ (𝐴 = {𝐶} → (𝐴 ⊆ {𝐵, 𝐶} ↔ {𝐶} ⊆ {𝐵, 𝐶}))
10	8, 9	mpbiri 168	. . 3 ⊢ (𝐴 = {𝐶} → 𝐴 ⊆ {𝐵, 𝐶})
11		eqimss 3302	. . 3 ⊢ (𝐴 = {𝐵, 𝐶} → 𝐴 ⊆ {𝐵, 𝐶})
12	10, 11	jaoi 728	. 2 ⊢ ((𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶}) → 𝐴 ⊆ {𝐵, 𝐶})
13	7, 12	jaoi 728	1 ⊢ (((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})) → 𝐴 ⊆ {𝐵, 𝐶})

Syntax hints:   → wi 4   ∨ wo 720   = wceq 1402   ⊆ wss 3220  ∅c0 3520  {csn 3708  {cpr 3709

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-dif 3222  df-un 3224  df-in 3226  df-ss 3233  df-nul 3521  df-pr 3715

This theorem is referenced by:  sstpr  3880  pwprss  3929