Theorem ss1oel2o 13189

Description: Any subset of ordinal one being an element of ordinal two is equivalent to excluded middle. A variation of exmid01 4121 which more directly illustrates the contrast with el2oss1o 13188. (Contributed by Jim Kingdon, 8-Aug-2022.)

Assertion

Ref	Expression
ss1oel2o	⊢ (EXMID ↔ ∀𝑥(𝑥 ⊆ 1o → 𝑥 ∈ 2o))

Proof of Theorem ss1oel2o

Step	Hyp	Ref	Expression
1		exmid01 4121	. 2 ⊢ (EXMID ↔ ∀𝑥(𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
2		df1o2 6326	. . . . 5 ⊢ 1o = {∅}
3	2	sseq2i 3124	. . . 4 ⊢ (𝑥 ⊆ 1o ↔ 𝑥 ⊆ {∅})
4		df2o2 6328	. . . . . 6 ⊢ 2o = {∅, {∅}}
5	4	eleq2i 2206	. . . . 5 ⊢ (𝑥 ∈ 2o ↔ 𝑥 ∈ {∅, {∅}})
6		vex 2689	. . . . . 6 ⊢ 𝑥 ∈ V
7	6	elpr 3548	. . . . 5 ⊢ (𝑥 ∈ {∅, {∅}} ↔ (𝑥 = ∅ ∨ 𝑥 = {∅}))
8	5, 7	bitri 183	. . . 4 ⊢ (𝑥 ∈ 2o ↔ (𝑥 = ∅ ∨ 𝑥 = {∅}))
9	3, 8	imbi12i 238	. . 3 ⊢ ((𝑥 ⊆ 1o → 𝑥 ∈ 2o) ↔ (𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
10	9	albii 1446	. 2 ⊢ (∀𝑥(𝑥 ⊆ 1o → 𝑥 ∈ 2o) ↔ ∀𝑥(𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
11	1, 10	bitr4i 186	1 ⊢ (EXMID ↔ ∀𝑥(𝑥 ⊆ 1o → 𝑥 ∈ 2o))

Syntax hints:   → wi 4   ↔ wb 104   ∨ wo 697  ∀wal 1329   = wceq 1331   ∈ wcel 1480   ⊆ wss 3071  ∅c0 3363  {csn 3527  {cpr 3528  EXMIDwem 4118  1_oc1o 6306  2_oc2o 6307

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 603  ax-in2 604  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121  ax-nul 4054

This theorem depends on definitions:  df-bi 116  df-dc 820  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-v 2688  df-dif 3073  df-un 3075  df-in 3077  df-ss 3084  df-nul 3364  df-sn 3533  df-pr 3534  df-exmid 4119  df-suc 4293  df-1o 6313  df-2o 6314

This theorem is referenced by: (None)