Theorem pwprss 3567

Description: The power set of an unordered pair. (Contributed by Jim Kingdon, 13-Aug-2018.)

Assertion

Ref	Expression
pwprss	⊢ ({∅, {A}} ∪ {{B}, {A, B}}) ⊆ 𝒫 {A, B}

Proof of Theorem pwprss

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		vex 2554	. . . . . 6 ⊢ x ∈ V
2	1	elpr 3385	. . . . 5 ⊢ (x ∈ {∅, {A}} ↔ (x = ∅ ∨ x = {A}))
3	1	elpr 3385	. . . . 5 ⊢ (x ∈ {{B}, {A, B}} ↔ (x = {B} ∨ x = {A, B}))
4	2, 3	orbi12i 680	. . . 4 ⊢ ((x ∈ {∅, {A}} ∨ x ∈ {{B}, {A, B}}) ↔ ((x = ∅ ∨ x = {A}) ∨ (x = {B} ∨ x = {A, B})))
5		ssprr 3518	. . . 4 ⊢ (((x = ∅ ∨ x = {A}) ∨ (x = {B} ∨ x = {A, B})) → x ⊆ {A, B})
6	4, 5	sylbi 114	. . 3 ⊢ ((x ∈ {∅, {A}} ∨ x ∈ {{B}, {A, B}}) → x ⊆ {A, B})
7		elun 3078	. . 3 ⊢ (x ∈ ({∅, {A}} ∪ {{B}, {A, B}}) ↔ (x ∈ {∅, {A}} ∨ x ∈ {{B}, {A, B}}))
8	1	elpw 3357	. . 3 ⊢ (x ∈ 𝒫 {A, B} ↔ x ⊆ {A, B})
9	6, 7, 8	3imtr4i 190	. 2 ⊢ (x ∈ ({∅, {A}} ∪ {{B}, {A, B}}) → x ∈ 𝒫 {A, B})
10	9	ssriv 2943	1 ⊢ ({∅, {A}} ∪ {{B}, {A, B}}) ⊆ 𝒫 {A, B}

Syntax hints:   ∨ wo 628   = wceq 1242   ∈ wcel 1390   ∪ cun 2909   ⊆ wss 2911  ∅c0 3218  𝒫 cpw 3351  {csn 3367  {cpr 3368

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bndl 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-dif 2914  df-un 2916  df-in 2918  df-ss 2925  df-nul 3219  df-pw 3353  df-sn 3373  df-pr 3374

This theorem is referenced by:  pwpwpw0ss  3569  ord3ex  3932