Theorem ssprr 3799

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 3500	. . . 4 ⊢ ∅ ⊆ {𝐵, 𝐶}
2		sseq1 3217	. . . 4 ⊢ (𝐴 = ∅ → (𝐴 ⊆ {𝐵, 𝐶} ↔ ∅ ⊆ {𝐵, 𝐶}))
3	1, 2	mpbiri 168	. . 3 ⊢ (𝐴 = ∅ → 𝐴 ⊆ {𝐵, 𝐶})
4		snsspr1 3783	. . . 4 ⊢ {𝐵} ⊆ {𝐵, 𝐶}
5		sseq1 3217	. . . 4 ⊢ (𝐴 = {𝐵} → (𝐴 ⊆ {𝐵, 𝐶} ↔ {𝐵} ⊆ {𝐵, 𝐶}))
6	4, 5	mpbiri 168	. . 3 ⊢ (𝐴 = {𝐵} → 𝐴 ⊆ {𝐵, 𝐶})
7	3, 6	jaoi 718	. 2 ⊢ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) → 𝐴 ⊆ {𝐵, 𝐶})
8		snsspr2 3784	. . . 4 ⊢ {𝐶} ⊆ {𝐵, 𝐶}
9		sseq1 3217	. . . 4 ⊢ (𝐴 = {𝐶} → (𝐴 ⊆ {𝐵, 𝐶} ↔ {𝐶} ⊆ {𝐵, 𝐶}))
10	8, 9	mpbiri 168	. . 3 ⊢ (𝐴 = {𝐶} → 𝐴 ⊆ {𝐵, 𝐶})
11		eqimss 3248	. . 3 ⊢ (𝐴 = {𝐵, 𝐶} → 𝐴 ⊆ {𝐵, 𝐶})
12	10, 11	jaoi 718	. 2 ⊢ ((𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶}) → 𝐴 ⊆ {𝐵, 𝐶})
13	7, 12	jaoi 718	1 ⊢ (((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})) → 𝐴 ⊆ {𝐵, 𝐶})

Syntax hints:   → wi 4   ∨ wo 710   = wceq 1373   ⊆ wss 3167  ∅c0 3461  {csn 3634  {cpr 3635

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 615  ax-in2 616  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-v 2775  df-dif 3169  df-un 3171  df-in 3173  df-ss 3180  df-nul 3462  df-pr 3641

This theorem is referenced by:  sstpr  3800  pwprss  3848