Theorem pw0 3727

Description: Compute the power set of the empty set. Theorem 89 of [Suppes] p. 47. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Andrew Salmon, 29-Jun-2011.)

Assertion

Ref	Expression
pw0	⊢ 𝒫 ∅ = {∅}

Proof of Theorem pw0

Step	Hyp	Ref	Expression
1		ss0b 3454	. . 3 ⊢ (𝑥 ⊆ ∅ ↔ 𝑥 = ∅)
2	1	abbii 2286	. 2 ⊢ {𝑥 ∣ 𝑥 ⊆ ∅} = {𝑥 ∣ 𝑥 = ∅}
3		df-pw 3568	. 2 ⊢ 𝒫 ∅ = {𝑥 ∣ 𝑥 ⊆ ∅}
4		df-sn 3589	. 2 ⊢ {∅} = {𝑥 ∣ 𝑥 = ∅}
5	2, 3, 4	3eqtr4i 2201	1 ⊢ 𝒫 ∅ = {∅}

Syntax hints:   = wceq 1348  {cab 2156   ⊆ wss 3121  ∅c0 3414  𝒫 cpw 3566  {csn 3583

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 609  ax-in2 610  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-dif 3123  df-in 3127  df-ss 3134  df-nul 3415  df-pw 3568  df-sn 3589

This theorem is referenced by:  p0ex  4174  sn0topon  12882  sn0cld  12931