Theorem pwpr 4870

Description: The power set of an unordered pair. (Contributed by NM, 1-May-2009.)

Assertion

Ref	Expression
pwpr	⊢ 𝒫 {𝐴, 𝐵} = ({∅, {𝐴}} ∪ {{𝐵}, {𝐴, 𝐵}})

Proof of Theorem pwpr

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sspr 4804	. . . 4 ⊢ (𝑥 ⊆ {𝐴, 𝐵} ↔ ((𝑥 = ∅ ∨ 𝑥 = {𝐴}) ∨ (𝑥 = {𝐵} ∨ 𝑥 = {𝐴, 𝐵})))
2		vex 3467	. . . . . 6 ⊢ 𝑥 ∈ V
3	2	elpr 4619	. . . . 5 ⊢ (𝑥 ∈ {∅, {𝐴}} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝐴}))
4	2	elpr 4619	. . . . 5 ⊢ (𝑥 ∈ {{𝐵}, {𝐴, 𝐵}} ↔ (𝑥 = {𝐵} ∨ 𝑥 = {𝐴, 𝐵}))
5	3, 4	orbi12i 927	. . . 4 ⊢ ((𝑥 ∈ {∅, {𝐴}} ∨ 𝑥 ∈ {{𝐵}, {𝐴, 𝐵}}) ↔ ((𝑥 = ∅ ∨ 𝑥 = {𝐴}) ∨ (𝑥 = {𝐵} ∨ 𝑥 = {𝐴, 𝐵})))
6	1, 5	bitr4i 281	. . 3 ⊢ (𝑥 ⊆ {𝐴, 𝐵} ↔ (𝑥 ∈ {∅, {𝐴}} ∨ 𝑥 ∈ {{𝐵}, {𝐴, 𝐵}}))
7		velpw 4572	. . 3 ⊢ (𝑥 ∈ 𝒫 {𝐴, 𝐵} ↔ 𝑥 ⊆ {𝐴, 𝐵})
8		elun 4115	. . 3 ⊢ (𝑥 ∈ ({∅, {𝐴}} ∪ {{𝐵}, {𝐴, 𝐵}}) ↔ (𝑥 ∈ {∅, {𝐴}} ∨ 𝑥 ∈ {{𝐵}, {𝐴, 𝐵}}))
9	6, 7, 8	3bitr4i 306	. 2 ⊢ (𝑥 ∈ 𝒫 {𝐴, 𝐵} ↔ 𝑥 ∈ ({∅, {𝐴}} ∪ {{𝐵}, {𝐴, 𝐵}}))
10	9	eqriv 2766	1 ⊢ 𝒫 {𝐴, 𝐵} = ({∅, {𝐴}} ∪ {{𝐵}, {𝐴, 𝐵}})

Syntax hints:   ∨ wo 860   = wceq 1567   ∈ wcel 2149   ∪ cun 3911   ⊆ wss 3913  ∅c0 4294  𝒫 cpw 4567  {csn 4594  {cpr 4596

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ral 3086  df-v 3465  df-dif 3916  df-un 3918  df-in 3920  df-ss 3930  df-nul 4295  df-pw 4569  df-sn 4595  df-pr 4597

This theorem is referenced by:  pwpwpw0  4872  ord3ex  5359  hash2pwpr  14512  pr2pwpr  14515  prsiga  34465  prsal  46923