Theorem pwpr 4838

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 4771	. . . 4 ⊢ (𝑥 ⊆ {𝐴, 𝐵} ↔ ((𝑥 = ∅ ∨ 𝑥 = {𝐴}) ∨ (𝑥 = {𝐵} ∨ 𝑥 = {𝐴, 𝐵})))
2		vex 3434	. . . . . 6 ⊢ 𝑥 ∈ V
3	2	elpr 4589	. . . . 5 ⊢ (𝑥 ∈ {∅, {𝐴}} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝐴}))
4	2	elpr 4589	. . . . 5 ⊢ (𝑥 ∈ {{𝐵}, {𝐴, 𝐵}} ↔ (𝑥 = {𝐵} ∨ 𝑥 = {𝐴, 𝐵}))
5	3, 4	orbi12i 911	. . . 4 ⊢ ((𝑥 ∈ {∅, {𝐴}} ∨ 𝑥 ∈ {{𝐵}, {𝐴, 𝐵}}) ↔ ((𝑥 = ∅ ∨ 𝑥 = {𝐴}) ∨ (𝑥 = {𝐵} ∨ 𝑥 = {𝐴, 𝐵})))
6	1, 5	bitr4i 277	. . 3 ⊢ (𝑥 ⊆ {𝐴, 𝐵} ↔ (𝑥 ∈ {∅, {𝐴}} ∨ 𝑥 ∈ {{𝐵}, {𝐴, 𝐵}}))
7		velpw 4543	. . 3 ⊢ (𝑥 ∈ 𝒫 {𝐴, 𝐵} ↔ 𝑥 ⊆ {𝐴, 𝐵})
8		elun 4087	. . 3 ⊢ (𝑥 ∈ ({∅, {𝐴}} ∪ {{𝐵}, {𝐴, 𝐵}}) ↔ (𝑥 ∈ {∅, {𝐴}} ∨ 𝑥 ∈ {{𝐵}, {𝐴, 𝐵}}))
9	6, 7, 8	3bitr4i 302	. 2 ⊢ (𝑥 ∈ 𝒫 {𝐴, 𝐵} ↔ 𝑥 ∈ ({∅, {𝐴}} ∪ {{𝐵}, {𝐴, 𝐵}}))
10	9	eqriv 2736	1 ⊢ 𝒫 {𝐴, 𝐵} = ({∅, {𝐴}} ∪ {{𝐵}, {𝐴, 𝐵}})

Syntax hints:   ∨ wo 843   = wceq 1541   ∈ wcel 2109   ∪ cun 3889   ⊆ wss 3891  ∅c0 4261  𝒫 cpw 4538  {csn 4566  {cpr 4568

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1801  ax-4 1815  ax-5 1916  ax-6 1974  ax-7 2014  ax-8 2111  ax-9 2119  ax-ext 2710

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1544  df-fal 1554  df-ex 1786  df-sb 2071  df-clab 2717  df-cleq 2731  df-clel 2817  df-ral 3070  df-v 3432  df-dif 3894  df-un 3896  df-in 3898  df-ss 3908  df-nul 4262  df-pw 4540  df-sn 4567  df-pr 4569

This theorem is referenced by:  pwpwpw0  4840  ord3ex  5313  hash2pwpr  14171  pr2pwpr  14174  prsiga  32078  prsal  43813