Theorem pwpw0 4834

Description: Compute the power set of the power set of the empty set. (See pw0 4833 for the power set of the empty set.) Theorem 90 of [Suppes] p. 48. Although this theorem is a special case of pwsn 4830, we have chosen to show a direct elementary proof. (Contributed by NM, 7-Aug-1994.) Avoid pwsn 4830. (Proof modification is discouraged.)

Assertion

Ref	Expression
pwpw0	⊢ 𝒫 {∅} = {∅, {∅}}

Proof of Theorem pwpw0

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-ss 3901	. . . . . . . . 9 ⊢ (𝑥 ⊆ {∅} ↔ ∀𝑦(𝑦 ∈ 𝑥 → 𝑦 ∈ {∅}))
2		velsn 4600	. . . . . . . . . . 11 ⊢ (𝑦 ∈ {∅} ↔ 𝑦 = ∅)
3	2	imbi2i 339	. . . . . . . . . 10 ⊢ ((𝑦 ∈ 𝑥 → 𝑦 ∈ {∅}) ↔ (𝑦 ∈ 𝑥 → 𝑦 = ∅))
4	3	albii 1852	. . . . . . . . 9 ⊢ (∀𝑦(𝑦 ∈ 𝑥 → 𝑦 ∈ {∅}) ↔ ∀𝑦(𝑦 ∈ 𝑥 → 𝑦 = ∅))
5	1, 4	bitri 278	. . . . . . . 8 ⊢ (𝑥 ⊆ {∅} ↔ ∀𝑦(𝑦 ∈ 𝑥 → 𝑦 = ∅))
6		neq0 4302	. . . . . . . . . 10 ⊢ (¬ 𝑥 = ∅ ↔ ∃𝑦 𝑦 ∈ 𝑥)
7		exintr 1927	. . . . . . . . . 10 ⊢ (∀𝑦(𝑦 ∈ 𝑥 → 𝑦 = ∅) → (∃𝑦 𝑦 ∈ 𝑥 → ∃𝑦(𝑦 ∈ 𝑥 ∧ 𝑦 = ∅)))
8	6, 7	biimtrid 245	. . . . . . . . 9 ⊢ (∀𝑦(𝑦 ∈ 𝑥 → 𝑦 = ∅) → (¬ 𝑥 = ∅ → ∃𝑦(𝑦 ∈ 𝑥 ∧ 𝑦 = ∅)))
9		exancom 1894	. . . . . . . . . . 11 ⊢ (∃𝑦(𝑦 ∈ 𝑥 ∧ 𝑦 = ∅) ↔ ∃𝑦(𝑦 = ∅ ∧ 𝑦 ∈ 𝑥))
10		dfclel 2840	. . . . . . . . . . 11 ⊢ (∅ ∈ 𝑥 ↔ ∃𝑦(𝑦 = ∅ ∧ 𝑦 ∈ 𝑥))
11	9, 10	bitr4i 281	. . . . . . . . . 10 ⊢ (∃𝑦(𝑦 ∈ 𝑥 ∧ 𝑦 = ∅) ↔ ∅ ∈ 𝑥)
12		snssi 4746	. . . . . . . . . 10 ⊢ (∅ ∈ 𝑥 → {∅} ⊆ 𝑥)
13	11, 12	sylbi 220	. . . . . . . . 9 ⊢ (∃𝑦(𝑦 ∈ 𝑥 ∧ 𝑦 = ∅) → {∅} ⊆ 𝑥)
14	8, 13	syl6 36	. . . . . . . 8 ⊢ (∀𝑦(𝑦 ∈ 𝑥 → 𝑦 = ∅) → (¬ 𝑥 = ∅ → {∅} ⊆ 𝑥))
15	5, 14	sylbi 220	. . . . . . 7 ⊢ (𝑥 ⊆ {∅} → (¬ 𝑥 = ∅ → {∅} ⊆ 𝑥))
16	15	anc2li 565	. . . . . 6 ⊢ (𝑥 ⊆ {∅} → (¬ 𝑥 = ∅ → (𝑥 ⊆ {∅} ∧ {∅} ⊆ 𝑥)))
17		eqss 3928	. . . . . 6 ⊢ (𝑥 = {∅} ↔ (𝑥 ⊆ {∅} ∧ {∅} ⊆ 𝑥))
18	16, 17	imbitrrdi 255	. . . . 5 ⊢ (𝑥 ⊆ {∅} → (¬ 𝑥 = ∅ → 𝑥 = {∅}))
19	18	orrd 877	. . . 4 ⊢ (𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅}))
20		0ss 4353	. . . . . 6 ⊢ ∅ ⊆ {∅}
21		sseq1 3938	. . . . . 6 ⊢ (𝑥 = ∅ → (𝑥 ⊆ {∅} ↔ ∅ ⊆ {∅}))
22	20, 21	mpbiri 261	. . . . 5 ⊢ (𝑥 = ∅ → 𝑥 ⊆ {∅})
23		eqimss 3971	. . . . 5 ⊢ (𝑥 = {∅} → 𝑥 ⊆ {∅})
24	22, 23	jaoi 871	. . . 4 ⊢ ((𝑥 = ∅ ∨ 𝑥 = {∅}) → 𝑥 ⊆ {∅})
25	19, 24	impbii 212	. . 3 ⊢ (𝑥 ⊆ {∅} ↔ (𝑥 = ∅ ∨ 𝑥 = {∅}))
26	25	abbii 2830	. 2 ⊢ {𝑥 ∣ 𝑥 ⊆ {∅}} = {𝑥 ∣ (𝑥 = ∅ ∨ 𝑥 = {∅})}
27		df-pw 4562	. 2 ⊢ 𝒫 {∅} = {𝑥 ∣ 𝑥 ⊆ {∅}}
28		dfpr2 4605	. 2 ⊢ {∅, {∅}} = {𝑥 ∣ (𝑥 = ∅ ∨ 𝑥 = {∅})}
29	26, 27, 28	3eqtr4i 2796	1 ⊢ 𝒫 {∅} = {∅, {∅}}

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∨ wo 861  ∀wal 1568   = wceq 1570  ∃wex 1812   ∈ wcel 2147  {cab 2741   ⊆ wss 3900  ∅c0 4281  𝒫 cpw 4560  {csn 4588  {cpr 4590

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-ss 3901  df-dif 4050  df-un 4154  df-nul 4282  df-pw 4562  df-sn 4589  df-pr 4591

This theorem is used by:  pp0ex  5353  pwdju1  10299  canthp1lem1  10767  r12  35823  rankeq1o  37123  ssoninhaus  37427