Theorem ss1oel2o 15484

Description: Any subset of ordinal one being an element of ordinal two is equivalent to excluded middle. A variation of exmid01 4227 which more directly illustrates the contrast with el2oss1o 6496. (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 4227	. 2 ⊢ (EXMID ↔ ∀𝑥(𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
2		df1o2 6482	. . . . 5 ⊢ 1o = {∅}
3	2	sseq2i 3206	. . . 4 ⊢ (𝑥 ⊆ 1o ↔ 𝑥 ⊆ {∅})
4		df2o2 6484	. . . . . 6 ⊢ 2o = {∅, {∅}}
5	4	eleq2i 2260	. . . . 5 ⊢ (𝑥 ∈ 2o ↔ 𝑥 ∈ {∅, {∅}})
6		vex 2763	. . . . . 6 ⊢ 𝑥 ∈ V
7	6	elpr 3639	. . . . 5 ⊢ (𝑥 ∈ {∅, {∅}} ↔ (𝑥 = ∅ ∨ 𝑥 = {∅}))
8	5, 7	bitri 184	. . . 4 ⊢ (𝑥 ∈ 2o ↔ (𝑥 = ∅ ∨ 𝑥 = {∅}))
9	3, 8	imbi12i 239	. . 3 ⊢ ((𝑥 ⊆ 1o → 𝑥 ∈ 2o) ↔ (𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
10	9	albii 1481	. 2 ⊢ (∀𝑥(𝑥 ⊆ 1o → 𝑥 ∈ 2o) ↔ ∀𝑥(𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
11	1, 10	bitr4i 187	1 ⊢ (EXMID ↔ ∀𝑥(𝑥 ⊆ 1o → 𝑥 ∈ 2o))

Syntax hints:   → wi 4   ↔ wb 105   ∨ wo 709  ∀wal 1362   = wceq 1364   ∈ wcel 2164   ⊆ wss 3153  ∅c0 3446  {csn 3618  {cpr 3619  EXMIDwem 4223  1_oc1o 6462  2_oc2o 6463

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 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175  ax-nul 4155

This theorem depends on definitions:  df-bi 117  df-dc 836  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-v 2762  df-dif 3155  df-un 3157  df-in 3159  df-ss 3166  df-nul 3447  df-sn 3624  df-pr 3625  df-exmid 4224  df-suc 4402  df-1o 6469  df-2o 6470

This theorem is referenced by: (None)