Theorem ss1oel2o 14026

Description: Any subset of ordinal one being an element of ordinal two is equivalent to excluded middle. A variation of exmid01 4184 which more directly illustrates the contrast with el2oss1o 6422. (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 4184	. 2 ⊢ (EXMID ↔ ∀𝑥(𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
2		df1o2 6408	. . . . 5 ⊢ 1o = {∅}
3	2	sseq2i 3174	. . . 4 ⊢ (𝑥 ⊆ 1o ↔ 𝑥 ⊆ {∅})
4		df2o2 6410	. . . . . 6 ⊢ 2o = {∅, {∅}}
5	4	eleq2i 2237	. . . . 5 ⊢ (𝑥 ∈ 2o ↔ 𝑥 ∈ {∅, {∅}})
6		vex 2733	. . . . . 6 ⊢ 𝑥 ∈ V
7	6	elpr 3604	. . . . 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 1463	. 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 703  ∀wal 1346   = wceq 1348   ∈ wcel 2141   ⊆ wss 3121  ∅c0 3414  {csn 3583  {cpr 3584  EXMIDwem 4180  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  ax-nul 4115

This theorem depends on definitions:  df-bi 116  df-dc 830  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-exmid 4181  df-suc 4356  df-1o 6395  df-2o 6396

This theorem is referenced by: (None)