Theorem ss1oel2o 16761

Description: Any subset of ordinal one being an element of ordinal two is equivalent to excluded middle. A variation of exmid01 4311 which more directly illustrates the contrast with el2oss1o 6676. (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 4311	. 2 ⊢ (EXMID ↔ ∀𝑥(𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
2		df1o2 6661	. . . . 5 ⊢ 1o = {∅}
3	2	sseq2i 3265	. . . 4 ⊢ (𝑥 ⊆ 1o ↔ 𝑥 ⊆ {∅})
4		df2o2 6663	. . . . . 6 ⊢ 2o = {∅, {∅}}
5	4	eleq2i 2299	. . . . 5 ⊢ (𝑥 ∈ 2o ↔ 𝑥 ∈ {∅, {∅}})
6		vex 2816	. . . . . 6 ⊢ 𝑥 ∈ V
7	6	elpr 3710	. . . . 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 1519	. 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 716  ∀wal 1396   = wceq 1398   ∈ wcel 2203   ⊆ wss 3211  ∅c0 3508  {csn 3689  {cpr 3690  EXMIDwem 4307  1_oc1o 6640  2_oc2o 6641

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 619  ax-in2 620  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2214  ax-nul 4236

This theorem depends on definitions:  df-bi 117  df-dc 843  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2219  df-cleq 2225  df-clel 2228  df-nfc 2373  df-v 2815  df-dif 3213  df-un 3215  df-in 3217  df-ss 3224  df-nul 3509  df-sn 3695  df-pr 3696  df-exmid 4308  df-suc 4492  df-1o 6647  df-2o 6648

This theorem is referenced by: (None)