Theorem pwpr 4903

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 4837	. . . 4 ⊢ (𝑥 ⊆ {𝐴, 𝐵} ↔ ((𝑥 = ∅ ∨ 𝑥 = {𝐴}) ∨ (𝑥 = {𝐵} ∨ 𝑥 = {𝐴, 𝐵})))
2		vex 3479	. . . . . 6 ⊢ 𝑥 ∈ V
3	2	elpr 4652	. . . . 5 ⊢ (𝑥 ∈ {∅, {𝐴}} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝐴}))
4	2	elpr 4652	. . . . 5 ⊢ (𝑥 ∈ {{𝐵}, {𝐴, 𝐵}} ↔ (𝑥 = {𝐵} ∨ 𝑥 = {𝐴, 𝐵}))
5	3, 4	orbi12i 914	. . . 4 ⊢ ((𝑥 ∈ {∅, {𝐴}} ∨ 𝑥 ∈ {{𝐵}, {𝐴, 𝐵}}) ↔ ((𝑥 = ∅ ∨ 𝑥 = {𝐴}) ∨ (𝑥 = {𝐵} ∨ 𝑥 = {𝐴, 𝐵})))
6	1, 5	bitr4i 278	. . 3 ⊢ (𝑥 ⊆ {𝐴, 𝐵} ↔ (𝑥 ∈ {∅, {𝐴}} ∨ 𝑥 ∈ {{𝐵}, {𝐴, 𝐵}}))
7		velpw 4608	. . 3 ⊢ (𝑥 ∈ 𝒫 {𝐴, 𝐵} ↔ 𝑥 ⊆ {𝐴, 𝐵})
8		elun 4149	. . 3 ⊢ (𝑥 ∈ ({∅, {𝐴}} ∪ {{𝐵}, {𝐴, 𝐵}}) ↔ (𝑥 ∈ {∅, {𝐴}} ∨ 𝑥 ∈ {{𝐵}, {𝐴, 𝐵}}))
9	6, 7, 8	3bitr4i 303	. 2 ⊢ (𝑥 ∈ 𝒫 {𝐴, 𝐵} ↔ 𝑥 ∈ ({∅, {𝐴}} ∪ {{𝐵}, {𝐴, 𝐵}}))
10	9	eqriv 2730	1 ⊢ 𝒫 {𝐴, 𝐵} = ({∅, {𝐴}} ∪ {{𝐵}, {𝐴, 𝐵}})

Syntax hints:   ∨ wo 846   = wceq 1542   ∈ wcel 2107   ∪ cun 3947   ⊆ wss 3949  ∅c0 4323  𝒫 cpw 4603  {csn 4629  {cpr 4631

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-tru 1545  df-fal 1555  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-ral 3063  df-v 3477  df-dif 3952  df-un 3954  df-in 3956  df-ss 3966  df-nul 4324  df-pw 4605  df-sn 4630  df-pr 4632

This theorem is referenced by:  pwpwpw0  4905  ord3ex  5386  hash2pwpr  14437  pr2pwpr  14440  prsiga  33129  prsal  45034