Theorem pw0 1882

Description: Compute the power set of the empty set. Theorem 89 of [Suppes] p. 47.

Assertion

Ref	Expression
pw0	⊢ ℘∅ = {∅}

Proof of Theorem pw0

Step	Hyp	Ref	Expression
1		df-pw 1799	. . . . 5 ⊢ ℘∅ = {x∣x ⊆ ∅}
2	1	cleqabi 1176	. . . 4 ⊢ (x ∈ ℘∅ ↔ x ⊆ ∅)
3		eqss 1516	. . . . 5 ⊢ (x = ∅ ↔ (x ⊆ ∅ ∧ ∅ ⊆ x))
4		0ss 1725	. . . . 5 ⊢ ∅ ⊆ x
5	3, 4	mpbiranr 548	. . . 4 ⊢ (x = ∅ ↔ x ⊆ ∅)
6	2, 5	bitr4 154	. . 3 ⊢ (x ∈ ℘∅ ↔ x = ∅)
7	6	biabri 1180	. 2 ⊢ ℘∅ = {x∣x = ∅}
8		df-sn 1811	. 2 ⊢ {∅} = {x∣x = ∅}
9	7, 8	eqtr4 1122	1 ⊢ ℘∅ = {∅}

Syntax hints:  {cab 1090   = wceq 1091   ∈ wcel 1092   ⊆ wss 1487  ∅c0 1707  ℘cpw 1798  {csn 1808

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-dif 1489  df-in 1491  df-ss 1492  df-nul 1708  df-pw 1799  df-sn 1811

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