Theorem pw0 3815

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 3531	. . 3 ⊢ (𝑥 ⊆ ∅ ↔ 𝑥 = ∅)
2	1	abbii 2345	. 2 ⊢ {𝑥 ∣ 𝑥 ⊆ ∅} = {𝑥 ∣ 𝑥 = ∅}
3		df-pw 3651	. 2 ⊢ 𝒫 ∅ = {𝑥 ∣ 𝑥 ⊆ ∅}
4		df-sn 3672	. 2 ⊢ {∅} = {𝑥 ∣ 𝑥 = ∅}
5	2, 3, 4	3eqtr4i 2260	1 ⊢ 𝒫 ∅ = {∅}

Syntax hints:   = wceq 1395  {cab 2215   ⊆ wss 3197  ∅c0 3491  𝒫 cpw 3649  {csn 3666

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 617  ax-in2 618  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2801  df-dif 3199  df-in 3203  df-ss 3210  df-nul 3492  df-pw 3651  df-sn 3672

This theorem is referenced by:  p0ex  4272  sn0topon  14770  sn0cld  14819  pw0ss  15891