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Theorem pwundifOLD 5477
Description: Obsolete proof of pwundif 4556 as of 26-Dec-2023. (Contributed by NM, 25-Mar-2007.) (Proof shortened by Thierry Arnoux, 20-Dec-2016.) (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
pwundifOLD 𝒫 (𝐴𝐵) = ((𝒫 (𝐴𝐵) ∖ 𝒫 𝐴) ∪ 𝒫 𝐴)

Proof of Theorem pwundifOLD
StepHypRef Expression
1 undif1 4406 . 2 ((𝒫 (𝐴𝐵) ∖ 𝒫 𝐴) ∪ 𝒫 𝐴) = (𝒫 (𝐴𝐵) ∪ 𝒫 𝐴)
2 pwunss 4550 . . . . 5 (𝒫 𝐴 ∪ 𝒫 𝐵) ⊆ 𝒫 (𝐴𝐵)
3 unss 4114 . . . . 5 ((𝒫 𝐴 ⊆ 𝒫 (𝐴𝐵) ∧ 𝒫 𝐵 ⊆ 𝒫 (𝐴𝐵)) ↔ (𝒫 𝐴 ∪ 𝒫 𝐵) ⊆ 𝒫 (𝐴𝐵))
42, 3mpbir 230 . . . 4 (𝒫 𝐴 ⊆ 𝒫 (𝐴𝐵) ∧ 𝒫 𝐵 ⊆ 𝒫 (𝐴𝐵))
54simpli 483 . . 3 𝒫 𝐴 ⊆ 𝒫 (𝐴𝐵)
6 ssequn2 4113 . . 3 (𝒫 𝐴 ⊆ 𝒫 (𝐴𝐵) ↔ (𝒫 (𝐴𝐵) ∪ 𝒫 𝐴) = 𝒫 (𝐴𝐵))
75, 6mpbi 229 . 2 (𝒫 (𝐴𝐵) ∪ 𝒫 𝐴) = 𝒫 (𝐴𝐵)
81, 7eqtr2i 2767 1 𝒫 (𝐴𝐵) = ((𝒫 (𝐴𝐵) ∖ 𝒫 𝐴) ∪ 𝒫 𝐴)
Colors of variables: wff setvar class
Syntax hints:  wa 395   = wceq 1539  cdif 3880  cun 3881  wss 3883  𝒫 cpw 4530
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-fal 1552  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-rab 3072  df-v 3424  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-pw 4532
This theorem is referenced by: (None)
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