Theorem el2oss1o 6422

Description: Being an element of ordinal two implies being a subset of ordinal one. The converse is equivalent to excluded middle by ss1oel2o 14026. (Contributed by Jim Kingdon, 8-Aug-2022.)

Assertion

Ref	Expression
el2oss1o	⊢ (𝐴 ∈ 2o → 𝐴 ⊆ 1o)

Proof of Theorem el2oss1o

Step	Hyp	Ref	Expression
1		elpri 3606	. . 3 ⊢ (𝐴 ∈ {∅, 1o} → (𝐴 = ∅ ∨ 𝐴 = 1o))
2		df2o3 6409	. . 3 ⊢ 2o = {∅, 1o}
3	1, 2	eleq2s 2265	. 2 ⊢ (𝐴 ∈ 2o → (𝐴 = ∅ ∨ 𝐴 = 1o))
4		0ss 3453	. . . 4 ⊢ ∅ ⊆ 1o
5		sseq1 3170	. . . 4 ⊢ (𝐴 = ∅ → (𝐴 ⊆ 1o ↔ ∅ ⊆ 1o))
6	4, 5	mpbiri 167	. . 3 ⊢ (𝐴 = ∅ → 𝐴 ⊆ 1o)
7		eqimss 3201	. . 3 ⊢ (𝐴 = 1o → 𝐴 ⊆ 1o)
8	6, 7	jaoi 711	. 2 ⊢ ((𝐴 = ∅ ∨ 𝐴 = 1o) → 𝐴 ⊆ 1o)
9	3, 8	syl 14	1 ⊢ (𝐴 ∈ 2o → 𝐴 ⊆ 1o)

Syntax hints:   → wi 4   ∨ wo 703   = wceq 1348   ∈ wcel 2141   ⊆ wss 3121  ∅c0 3414  {cpr 3584  1_oc1o 6388  2_oc2o 6389

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 609  ax-in2 610  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-dif 3123  df-un 3125  df-in 3127  df-ss 3134  df-nul 3415  df-sn 3589  df-pr 3590  df-suc 4356  df-1o 6395  df-2o 6396

This theorem is referenced by:  nnnninfeq2  7105  nninfwlpoimlemg  7151  nninfsellemsuc  14045