Theorem ss1oel2o 16313

Description: Any subset of ordinal one being an element of ordinal two is equivalent to excluded middle. A variation of exmid01 4281 which more directly illustrates the contrast with el2oss1o 6587. (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 4281	. 2 ⊢ (EXMID ↔ ∀𝑥(𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
2		df1o2 6573	. . . . 5 ⊢ 1o = {∅}
3	2	sseq2i 3251	. . . 4 ⊢ (𝑥 ⊆ 1o ↔ 𝑥 ⊆ {∅})
4		df2o2 6575	. . . . . 6 ⊢ 2o = {∅, {∅}}
5	4	eleq2i 2296	. . . . 5 ⊢ (𝑥 ∈ 2o ↔ 𝑥 ∈ {∅, {∅}})
6		vex 2802	. . . . . 6 ⊢ 𝑥 ∈ V
7	6	elpr 3687	. . . . 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 1516	. 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 713  ∀wal 1393   = wceq 1395   ∈ wcel 2200   ⊆ wss 3197  ∅c0 3491  {csn 3666  {cpr 3667  EXMIDwem 4277  1_oc1o 6553  2_oc2o 6554

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 617  ax-in2 618  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211  ax-nul 4209

This theorem depends on definitions:  df-bi 117  df-dc 840  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2801  df-dif 3199  df-un 3201  df-in 3203  df-ss 3210  df-nul 3492  df-sn 3672  df-pr 3673  df-exmid 4278  df-suc 4461  df-1o 6560  df-2o 6561

This theorem is referenced by: (None)