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Theorem pwpw0 4774
Description: Compute the power set of the power set of the empty set. (See pw0 4773 for the power set of the empty set.) Theorem 90 of [Suppes] p. 48. Although this theorem is a special case of pwsn 4860, we have chosen to show a direct elementary proof. (Contributed by NM, 7-Aug-1994.)
Assertion
Ref Expression
pwpw0 𝒫 {∅} = {∅, {∅}}

Proof of Theorem pwpw0
Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 df-ss 3916 . . . . . . . . 9 (𝑥 ⊆ {∅} ↔ ∀𝑦(𝑦 ∈ 𝑥 → 𝑦 ∈ {∅}))
2 velsn 4600 . . . . . . . . . . 11 (𝑦 ∈ {∅} ↔ 𝑦 = ∅)
32imbi2i 339 . . . . . . . . . 10 ((𝑦 ∈ 𝑥 → 𝑦 ∈ {∅}) ↔ (𝑦 ∈ 𝑥 → 𝑦 = ∅))
43albii 1852 . . . . . . . . 9 (∀𝑦(𝑦 ∈ 𝑥 → 𝑦 ∈ {∅}) ↔ ∀𝑦(𝑦 ∈ 𝑥 → 𝑦 = ∅))
51, 4bitri 278 . . . . . . . 8 (𝑥 ⊆ {∅} ↔ ∀𝑦(𝑦 ∈ 𝑥 → 𝑦 = ∅))
6 neq0 4299 . . . . . . . . . 10 (¬ 𝑥 = ∅ ↔ ∃𝑦 𝑦 ∈ 𝑥)
7 exintr 1925 . . . . . . . . . 10 (∀𝑦(𝑦 ∈ 𝑥 → 𝑦 = ∅) → (∃𝑦 𝑦 ∈ 𝑥 → ∃𝑦(𝑦 ∈ 𝑥 ∧ 𝑦 = ∅)))
86, 7biimtrid 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 2837 . . . . . . . . . . 11 (∅ ∈ 𝑥 ↔ ∃𝑦(𝑦 = ∅ ∧ 𝑦 ∈ 𝑥))
119, 10bitr4i 281 . . . . . . . . . 10 (∃𝑦(𝑦 ∈ 𝑥 ∧ 𝑦 = ∅) ↔ ∅ ∈ 𝑥)
12 snssi 4746 . . . . . . . . . 10 (∅ ∈ 𝑥 → {∅} ⊆ 𝑥)
1311, 12sylbi 220 . . . . . . . . 9 (∃𝑦(𝑦 ∈ 𝑥 ∧ 𝑦 = ∅) → {∅} ⊆ 𝑥)
148, 13syl6 36 . . . . . . . 8 (∀𝑦(𝑦 ∈ 𝑥 → 𝑦 = ∅) → (¬ 𝑥 = ∅ → {∅} ⊆ 𝑥))
155, 14sylbi 220 . . . . . . 7 (𝑥 ⊆ {∅} → (¬ 𝑥 = ∅ → {∅} ⊆ 𝑥))
1615anc2li 565 . . . . . 6 (𝑥 ⊆ {∅} → (¬ 𝑥 = ∅ → (𝑥 ⊆ {∅} ∧ {∅} ⊆ 𝑥)))
17 eqss 3946 . . . . . 6 (𝑥 = {∅} ↔ (𝑥 ⊆ {∅} ∧ {∅} ⊆ 𝑥))
1816, 17imbitrrdi 255 . . . . 5 (𝑥 ⊆ {∅} → (¬ 𝑥 = ∅ → 𝑥 = {∅}))
1918orrd 877 . . . 4 (𝑥 ⊆ {∅} → (𝑥 = ∅ ∨ 𝑥 = {∅}))
20 0ss 4350 . . . . . 6 ∅ ⊆ {∅}
21 sseq1 3956 . . . . . 6 (𝑥 = ∅ → (𝑥 ⊆ {∅} ↔ ∅ ⊆ {∅}))
2220, 21mpbiri 261 . . . . 5 (𝑥 = ∅ → 𝑥 ⊆ {∅})
23 eqimss 3989 . . . . 5 (𝑥 = {∅} → 𝑥 ⊆ {∅})
2422, 23jaoi 871 . . . 4 ((𝑥 = ∅ ∨ 𝑥 = {∅}) → 𝑥 ⊆ {∅})
2519, 24impbii 212 . . 3 (𝑥 ⊆ {∅} ↔ (𝑥 = ∅ ∨ 𝑥 = {∅}))
2625abbii 2828 . 2 {𝑥 ∣ 𝑥 ⊆ {∅}} = {𝑥 ∣ (𝑥 = ∅ ∨ 𝑥 = {∅})}
27 df-pw 4559 . 2 𝒫 {∅} = {𝑥 ∣ 𝑥 ⊆ {∅}}
28 dfpr2 4605 . 2 {∅, {∅}} = {𝑥 ∣ (𝑥 = ∅ ∨ 𝑥 = {∅})}
2926, 27, 283eqtr4i 2794 1 𝒫 {∅} = {∅, {∅}}
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∨ wo 861  ∀wal 1568   = wceq 1570  ∃wex 1812   ∈ wcel 2145  {cab 2739   ⊆ wss 3899  ∅c0 4279  𝒫 cpw 4557  {csn 4584  {cpr 4586
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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733
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 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-v 3453  df-dif 3902  df-un 3904  df-ss 3916  df-nul 4280  df-pw 4559  df-sn 4585  df-pr 4587
This theorem is used by:  pp0ex  5348  pwdju1  10250  canthp1lem1  10718  r12  35705  rankeq1o  36902  ssoninhaus  37206
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