Theorem exmidsssn 4181

Description: Excluded middle is equivalent to the biconditionalized version of sssnr 3733 for sets. (Contributed by Jim Kingdon, 5-Mar-2023.)

Assertion

Ref	Expression
exmidsssn	⊢ (EXMID ↔ ∀𝑥∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))

Distinct variable group:   𝑥,𝑦

Proof of Theorem exmidsssn

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		0ss 3447	. . . . . . 7 ⊢ ∅ ⊆ {𝑦}
2		sseq1 3165	. . . . . . 7 ⊢ (𝑥 = ∅ → (𝑥 ⊆ {𝑦} ↔ ∅ ⊆ {𝑦}))
3	1, 2	mpbiri 167	. . . . . 6 ⊢ (𝑥 = ∅ → 𝑥 ⊆ {𝑦})
4	3	adantl 275	. . . . 5 ⊢ ((EXMID ∧ 𝑥 = ∅) → 𝑥 ⊆ {𝑦})
5		simpr 109	. . . . . 6 ⊢ ((EXMID ∧ 𝑥 = ∅) → 𝑥 = ∅)
6	5	orcd 723	. . . . 5 ⊢ ((EXMID ∧ 𝑥 = ∅) → (𝑥 = ∅ ∨ 𝑥 = {𝑦}))
7	4, 6	2thd 174	. . . 4 ⊢ ((EXMID ∧ 𝑥 = ∅) → (𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
8		sssnm 3734	. . . . . 6 ⊢ (∃𝑧 𝑧 ∈ 𝑥 → (𝑥 ⊆ {𝑦} ↔ 𝑥 = {𝑦}))
9		neq0r 3423	. . . . . . 7 ⊢ (∃𝑧 𝑧 ∈ 𝑥 → ¬ 𝑥 = ∅)
10		biorf 734	. . . . . . 7 ⊢ (¬ 𝑥 = ∅ → (𝑥 = {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
11	9, 10	syl 14	. . . . . 6 ⊢ (∃𝑧 𝑧 ∈ 𝑥 → (𝑥 = {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
12	8, 11	bitrd 187	. . . . 5 ⊢ (∃𝑧 𝑧 ∈ 𝑥 → (𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
13	12	adantl 275	. . . 4 ⊢ ((EXMID ∧ ∃𝑧 𝑧 ∈ 𝑥) → (𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
14		exmidn0m 4180	. . . . . 6 ⊢ (EXMID ↔ ∀𝑥(𝑥 = ∅ ∨ ∃𝑧 𝑧 ∈ 𝑥))
15	14	biimpi 119	. . . . 5 ⊢ (EXMID → ∀𝑥(𝑥 = ∅ ∨ ∃𝑧 𝑧 ∈ 𝑥))
16	15	19.21bi 1546	. . . 4 ⊢ (EXMID → (𝑥 = ∅ ∨ ∃𝑧 𝑧 ∈ 𝑥))
17	7, 13, 16	mpjaodan 788	. . 3 ⊢ (EXMID → (𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
18	17	alrimivv 1863	. 2 ⊢ (EXMID → ∀𝑥∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
19		0ex 4109	. . . . . 6 ⊢ ∅ ∈ V
20		sneq 3587	. . . . . . . 8 ⊢ (𝑦 = ∅ → {𝑦} = {∅})
21	20	sseq2d 3172	. . . . . . 7 ⊢ (𝑦 = ∅ → (𝑥 ⊆ {𝑦} ↔ 𝑥 ⊆ {∅}))
22	20	eqeq2d 2177	. . . . . . . 8 ⊢ (𝑦 = ∅ → (𝑥 = {𝑦} ↔ 𝑥 = {∅}))
23	22	orbi2d 780	. . . . . . 7 ⊢ (𝑦 = ∅ → ((𝑥 = ∅ ∨ 𝑥 = {𝑦}) ↔ (𝑥 = ∅ ∨ 𝑥 = {∅})))
24	21, 23	bibi12d 234	. . . . . 6 ⊢ (𝑦 = ∅ → ((𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})) ↔ (𝑥 ⊆ {∅} ↔ (𝑥 = ∅ ∨ 𝑥 = {∅}))))
25	19, 24	spcv 2820	. . . . 5 ⊢ (∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})) → (𝑥 ⊆ {∅} ↔ (𝑥 = ∅ ∨ 𝑥 = {∅})))
26	25	biimpd 143	. . . 4 ⊢ (∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})) → (𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
27	26	alimi 1443	. . 3 ⊢ (∀𝑥∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})) → ∀𝑥(𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
28		exmid01 4177	. . 3 ⊢ (EXMID ↔ ∀𝑥(𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
29	27, 28	sylibr 133	. 2 ⊢ (∀𝑥∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})) → EXMID)
30	18, 29	impbii 125	1 ⊢ (EXMID ↔ ∀𝑥∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   ↔ wb 104   ∨ wo 698  ∀wal 1341   = wceq 1343  ∃wex 1480   ⊆ wss 3116  ∅c0 3409  {csn 3576  EXMIDwem 4173

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 604  ax-in2 605  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-14 2139  ax-ext 2147  ax-sep 4100  ax-nul 4108  ax-pow 4153

This theorem depends on definitions:  df-bi 116  df-dc 825  df-tru 1346  df-fal 1349  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-ne 2337  df-rab 2453  df-v 2728  df-dif 3118  df-in 3122  df-ss 3129  df-nul 3410  df-pw 3561  df-sn 3582  df-exmid 4174

This theorem is referenced by:  exmidsssnc  4182