Theorem exmidsssn 4257

Description: Excluded middle is equivalent to the biconditionalized version of sssnr 3802 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 3503	. . . . . . 7 ⊢ ∅ ⊆ {𝑦}
2		sseq1 3220	. . . . . . 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 735	. . . . 5 ⊢ ((EXMID ∧ 𝑥 = ∅) → (𝑥 = ∅ ∨ 𝑥 = {𝑦}))
7	4, 6	2thd 175	. . . 4 ⊢ ((EXMID ∧ 𝑥 = ∅) → (𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
8		sssnm 3803	. . . . . 6 ⊢ (∃𝑧 𝑧 ∈ 𝑥 → (𝑥 ⊆ {𝑦} ↔ 𝑥 = {𝑦}))
9		neq0r 3479	. . . . . . 7 ⊢ (∃𝑧 𝑧 ∈ 𝑥 → ¬ 𝑥 = ∅)
10		biorf 746	. . . . . . 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 4256	. . . . . 6 ⊢ (EXMID ↔ ∀𝑥(𝑥 = ∅ ∨ ∃𝑧 𝑧 ∈ 𝑥))
15	14	biimpi 120	. . . . 5 ⊢ (EXMID → ∀𝑥(𝑥 = ∅ ∨ ∃𝑧 𝑧 ∈ 𝑥))
16	15	19.21bi 1582	. . . 4 ⊢ (EXMID → (𝑥 = ∅ ∨ ∃𝑧 𝑧 ∈ 𝑥))
17	7, 13, 16	mpjaodan 800	. . 3 ⊢ (EXMID → (𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
18	17	alrimivv 1899	. 2 ⊢ (EXMID → ∀𝑥∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})))
19		0ex 4182	. . . . . 6 ⊢ ∅ ∈ V
20		sneq 3649	. . . . . . . 8 ⊢ (𝑦 = ∅ → {𝑦} = {∅})
21	20	sseq2d 3227	. . . . . . 7 ⊢ (𝑦 = ∅ → (𝑥 ⊆ {𝑦} ↔ 𝑥 ⊆ {∅}))
22	20	eqeq2d 2218	. . . . . . . 8 ⊢ (𝑦 = ∅ → (𝑥 = {𝑦} ↔ 𝑥 = {∅}))
23	22	orbi2d 792	. . . . . . 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 2871	. . . . 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 1479	. . 3 ⊢ (∀𝑥∀𝑦(𝑥 ⊆ {𝑦} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝑦})) → ∀𝑥(𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅})))
28		exmid01 4253	. . 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 710  ∀wal 1371   = wceq 1373  ∃wex 1516   ⊆ wss 3170  ∅c0 3464  {csn 3638  EXMIDwem 4249

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 615  ax-in2 616  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-14 2180  ax-ext 2188  ax-sep 4173  ax-nul 4181  ax-pow 4229

This theorem depends on definitions:  df-bi 117  df-dc 837  df-tru 1376  df-fal 1379  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-ne 2378  df-rab 2494  df-v 2775  df-dif 3172  df-in 3176  df-ss 3183  df-nul 3465  df-pw 3623  df-sn 3644  df-exmid 4250

This theorem is referenced by:  exmidsssnc  4258