Theorem pw0 3825

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 3536	. . 3 ⊢ (𝑥 ⊆ ∅ ↔ 𝑥 = ∅)
2	1	abbii 2347	. 2 ⊢ {𝑥 ∣ 𝑥 ⊆ ∅} = {𝑥 ∣ 𝑥 = ∅}
3		df-pw 3658	. 2 ⊢ 𝒫 ∅ = {𝑥 ∣ 𝑥 ⊆ ∅}
4		df-sn 3679	. 2 ⊢ {∅} = {𝑥 ∣ 𝑥 = ∅}
5	2, 3, 4	3eqtr4i 2262	1 ⊢ 𝒫 ∅ = {∅}

Syntax hints:   = wceq 1398  {cab 2217   ⊆ wss 3201  ∅c0 3496  𝒫 cpw 3656  {csn 3673

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 619  ax-in2 620  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-v 2805  df-dif 3203  df-in 3207  df-ss 3214  df-nul 3497  df-pw 3658  df-sn 3679

This theorem is referenced by:  p0ex  4284  sn0topon  14882  sn0cld  14931  pw0ss  16007