Theorem exmidsssn 4285

Description: Excluded middle is equivalent to the biconditionalized version of sssnr 3830 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 3530	. . . . . . 7 ⊢ ∅ ⊆ {𝑦}
2		sseq1 3247	. . . . . . 7 ⊢ (𝑥 = ∅ → (𝑥 ⊆ {𝑦} ↔ ∅ ⊆ {𝑦}))
3	1, 2	mpbiri 168	. . . . . 6 ⊢ (𝑥 = ∅ → 𝑥 ⊆ {𝑦})
4	3	adantl 277	. . . . 5 ⊢ ((EXMID ∧ 𝑥 = ∅) → 𝑥 ⊆ {𝑦})
5		simpr 110	. . . . . 6 ⊢ ((EXMID ∧ 𝑥 = ∅) → 𝑥 = ∅)
6	5	orcd 738	. . . . 5 ⊢ ((EXMID ∧ 𝑥 = ∅) → (𝑥 = ∅ ∨ 𝑥 = {𝑦}))
7	4, 6	2thd 175	. . . 4 ⊢ ((EXMID ∧ 𝑥 = ∅) → (𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
8		sssnm 3831	. . . . . 6 ⊢ (∃𝑧 𝑧 ∈ 𝑥 → (𝑥 ⊆ {𝑦} ↔ 𝑥 = {𝑦}))
9		neq0r 3506	. . . . . . 7 ⊢ (∃𝑧 𝑧 ∈ 𝑥 → ¬ 𝑥 = ∅)
10		biorf 749	. . . . . . 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 188	. . . . 5 ⊢ (∃𝑧 𝑧 ∈ 𝑥 → (𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
13	12	adantl 277	. . . 4 ⊢ ((EXMID ∧ ∃𝑧 𝑧 ∈ 𝑥) → (𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
14		exmidn0m 4284	. . . . . 6 ⊢ (EXMID ↔ ∀𝑥(𝑥 = ∅ ∨ ∃𝑧 𝑧 ∈ 𝑥))
15	14	biimpi 120	. . . . 5 ⊢ (EXMID → ∀𝑥(𝑥 = ∅ ∨ ∃𝑧 𝑧 ∈ 𝑥))
16	15	19.21bi 1604	. . . 4 ⊢ (EXMID → (𝑥 = ∅ ∨ ∃𝑧 𝑧 ∈ 𝑥))
17	7, 13, 16	mpjaodan 803	. . 3 ⊢ (EXMID → (𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
18	17	alrimivv 1921	. 2 ⊢ (EXMID → ∀𝑥∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
19		0ex 4210	. . . . . 6 ⊢ ∅ ∈ V
20		sneq 3677	. . . . . . . 8 ⊢ (𝑦 = ∅ → {𝑦} = {∅})
21	20	sseq2d 3254	. . . . . . 7 ⊢ (𝑦 = ∅ → (𝑥 ⊆ {𝑦} ↔ 𝑥 ⊆ {∅}))
22	20	eqeq2d 2241	. . . . . . . 8 ⊢ (𝑦 = ∅ → (𝑥 = {𝑦} ↔ 𝑥 = {∅}))
23	22	orbi2d 795	. . . . . . 7 ⊢ (𝑦 = ∅ → ((𝑥 = ∅ ∨ 𝑥 = {𝑦}) ↔ (𝑥 = ∅ ∨ 𝑥 = {∅})))
24	21, 23	bibi12d 235	. . . . . 6 ⊢ (𝑦 = ∅ → ((𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})) ↔ (𝑥 ⊆ {∅} ↔ (𝑥 = ∅ ∨ 𝑥 = {∅}))))
25	19, 24	spcv 2897	. . . . 5 ⊢ (∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})) → (𝑥 ⊆ {∅} ↔ (𝑥 = ∅ ∨ 𝑥 = {∅})))
26	25	biimpd 144	. . . 4 ⊢ (∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})) → (𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
27	26	alimi 1501	. . 3 ⊢ (∀𝑥∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})) → ∀𝑥(𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
28		exmid01 4281	. . 3 ⊢ (EXMID ↔ ∀𝑥(𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
29	27, 28	sylibr 134	. 2 ⊢ (∀𝑥∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})) → EXMID)
30	18, 29	impbii 126	1 ⊢ (EXMID ↔ ∀𝑥∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   ↔ wb 105   ∨ wo 713  ∀wal 1393   = wceq 1395  ∃wex 1538   ⊆ wss 3197  ∅c0 3491  {csn 3666  EXMIDwem 4277

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-14 2203  ax-ext 2211  ax-sep 4201  ax-nul 4209  ax-pow 4257

This theorem depends on definitions:  df-bi 117  df-dc 840  df-tru 1398  df-fal 1401  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-ne 2401  df-rab 2517  df-v 2801  df-dif 3199  df-in 3203  df-ss 3210  df-nul 3492  df-pw 3651  df-sn 3672  df-exmid 4278

This theorem is referenced by:  exmidsssnc  4286