Theorem pw0 3558

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 3304	. . 3 ⊢ (𝑥 ⊆ ∅ ↔ 𝑥 = ∅)
2	1	abbii 2198	. 2 ⊢ {𝑥 ∣ 𝑥 ⊆ ∅} = {𝑥 ∣ 𝑥 = ∅}
3		df-pw 3408	. 2 ⊢ 𝒫 ∅ = {𝑥 ∣ 𝑥 ⊆ ∅}
4		df-sn 3428	. 2 ⊢ {∅} = {𝑥 ∣ 𝑥 = ∅}
5	2, 3, 4	3eqtr4i 2113	1 ⊢ 𝒫 ∅ = {∅}

Syntax hints:   = wceq 1285  {cab 2069   ⊆ wss 2984  ∅c0 3269  𝒫 cpw 3406  {csn 3422

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 577  ax-in2 578  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2065

This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1688  df-clab 2070  df-cleq 2076  df-clel 2079  df-nfc 2212  df-v 2614  df-dif 2986  df-in 2990  df-ss 2997  df-nul 3270  df-pw 3408  df-sn 3428

This theorem is referenced by:  p0ex  3986