Theorem pwpr 3883

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

Assertion

Ref	Expression
pwpr	⊢ ℘{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 3869	. . . 4 ⊢ (x ⊆ {A, B} ↔ ((x = ∅ ∨ x = {A}) ∨ (x = {B} ∨ x = {A, B})))
2		vex 2862	. . . . . 6 ⊢ x ∈ V
3	2	elpr 3751	. . . . 5 ⊢ (x ∈ {∅, {A}} ↔ (x = ∅ ∨ x = {A}))
4	2	elpr 3751	. . . . 5 ⊢ (x ∈ {{B}, {A, B}} ↔ (x = {B} ∨ x = {A, B}))
5	3, 4	orbi12i 507	. . . 4 ⊢ ((x ∈ {∅, {A}} ∨ x ∈ {{B}, {A, B}}) ↔ ((x = ∅ ∨ x = {A}) ∨ (x = {B} ∨ x = {A, B})))
6	1, 5	bitr4i 243	. . 3 ⊢ (x ⊆ {A, B} ↔ (x ∈ {∅, {A}} ∨ x ∈ {{B}, {A, B}}))
7	2	elpw 3728	. . 3 ⊢ (x ∈ ℘{A, B} ↔ x ⊆ {A, B})
8		elun 3220	. . 3 ⊢ (x ∈ ({∅, {A}} ∪ {{B}, {A, B}}) ↔ (x ∈ {∅, {A}} ∨ x ∈ {{B}, {A, B}}))
9	6, 7, 8	3bitr4i 268	. 2 ⊢ (x ∈ ℘{A, B} ↔ x ∈ ({∅, {A}} ∪ {{B}, {A, B}}))
10	9	eqriv 2350	1 ⊢ ℘{A, B} = ({∅, {A}} ∪ {{B}, {A, B}})

Syntax hints:   ∨ wo 357   = wceq 1642   ∈ wcel 1710   ∪ cun 3207   ⊆ wss 3257  ∅c0 3550  ℘cpw 3722  {csn 3737  {cpr 3738

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-ne 2518  df-ral 2619  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-un 3214  df-dif 3215  df-ss 3259  df-nul 3551  df-pw 3724  df-sn 3741  df-pr 3742

This theorem is referenced by:  pwpwpw0  3885