Theorem pwpr 4833

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 4795	. . . 4 ⊢ (𝑥 ⊆ {𝐴, 𝐵} ↔ ((𝑥 = ∅ ∨ 𝑥 = {𝐴}) ∨ (𝑥 = {𝐵} ∨ 𝑥 = {𝐴, 𝐵})))
2		vex 3458	. . . . . 6 ⊢ 𝑥 ∈ V
3	2	elpr 4609	. . . . 5 ⊢ (𝑥 ∈ {∅, {𝐴}} ↔ (𝑥 = ∅ ∨ 𝑥 = {𝐴}))
4	2	elpr 4609	. . . . 5 ⊢ (𝑥 ∈ {{𝐵}, {𝐴, 𝐵}} ↔ (𝑥 = {𝐵} ∨ 𝑥 = {𝐴, 𝐵}))
5	3, 4	orbi12i 928	. . . 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 4565	. . 3 ⊢ (𝑥 ∈ 𝒫 {𝐴, 𝐵} ↔ 𝑥 ⊆ {𝐴, 𝐵})
8		elun 4155	. . 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 2760	1 ⊢ 𝒫 {𝐴, 𝐵} = ({∅, {𝐴}} ∪ {{𝐵}, {𝐴, 𝐵}})

This proof depends on syntax axioms:   ∨ wo 861   = wceq 1570   ∈ wcel 2147   ⊆ wss 3900   ∪ cun 4152  ∅c0 4281  𝒫 cpw 4560  {csn 4588  {cpr 4590

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-pw 4562  df-sn 4589  df-pr 4591

This theorem is used by:  pwpwpw0  4837  ord3ex  5356  hash2pwpr  14662  pr2pwpr  14665  prsiga  34874  prsal  47601