Theorem pwpr 4566

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 4497	. . . 4 ⊢ (𝑥 ⊆ {𝐴, 𝐵} ↔ ((𝑥 = ∅ ∨ 𝑥 = {𝐴}) ∨ (𝑥 = {𝐵} ∨ 𝑥 = {𝐴, 𝐵})))
2		vex 3352	. . . . . 6 ⊢ 𝑥 ∈ V
3	2	elpr 4336	. . . . 5 ⊢ (𝑥 ∈ {∅, {𝐴}} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝐴}))
4	2	elpr 4336	. . . . 5 ⊢ (𝑥 ∈ {{𝐵}, {𝐴, 𝐵}} ↔ (𝑥 = {𝐵} ∨ 𝑥 = {𝐴, 𝐵}))
5	3, 4	orbi12i 879	. . . 4 ⊢ ((𝑥 ∈ {∅, {𝐴}} ∨ 𝑥 ∈ {{𝐵}, {𝐴, 𝐵}}) ↔ ((𝑥 = ∅ ∨ 𝑥 = {𝐴}) ∨ (𝑥 = {𝐵} ∨ 𝑥 = {𝐴, 𝐵})))
6	1, 5	bitr4i 267	. . 3 ⊢ (𝑥 ⊆ {𝐴, 𝐵} ↔ (𝑥 ∈ {∅, {𝐴}} ∨ 𝑥 ∈ {{𝐵}, {𝐴, 𝐵}}))
7		selpw 4302	. . 3 ⊢ (𝑥 ∈ 𝒫 {𝐴, 𝐵} ↔ 𝑥 ⊆ {𝐴, 𝐵})
8		elun 3902	. . 3 ⊢ (𝑥 ∈ ({∅, {𝐴}} ∪ {{𝐵}, {𝐴, 𝐵}}) ↔ (𝑥 ∈ {∅, {𝐴}} ∨ 𝑥 ∈ {{𝐵}, {𝐴, 𝐵}}))
9	6, 7, 8	3bitr4i 292	. 2 ⊢ (𝑥 ∈ 𝒫 {𝐴, 𝐵} ↔ 𝑥 ∈ ({∅, {𝐴}} ∪ {{𝐵}, {𝐴, 𝐵}}))
10	9	eqriv 2767	1 ⊢ 𝒫 {𝐴, 𝐵} = ({∅, {𝐴}} ∪ {{𝐵}, {𝐴, 𝐵}})

Syntax hints:   ∨ wo 826   = wceq 1630   ∈ wcel 2144   ∪ cun 3719   ⊆ wss 3721  ∅c0 4061  𝒫 cpw 4295  {csn 4314  {cpr 4316

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1869  ax-4 1884  ax-5 1990  ax-6 2056  ax-7 2092  ax-9 2153  ax-10 2173  ax-11 2189  ax-12 2202  ax-13 2407  ax-ext 2750

This theorem depends on definitions:  df-bi 197  df-an 383  df-or 827  df-3an 1072  df-tru 1633  df-ex 1852  df-nf 1857  df-sb 2049  df-clab 2757  df-cleq 2763  df-clel 2766  df-nfc 2901  df-ral 3065  df-v 3351  df-dif 3724  df-un 3726  df-in 3728  df-ss 3735  df-nul 4062  df-pw 4297  df-sn 4315  df-pr 4317

This theorem is referenced by:  pwpwpw0  4568  ord3ex  4984  hash2pwpr  13459  pr2pwpr  13462  prsiga  30528  prsal  41049