Theorem pwundif 4592

Description: Break up the power class of a union into a union of smaller classes. (Contributed by NM, 25-Mar-2007.) (Proof shortened by Thierry Arnoux, 20-Dec-2016.) Remove use of ax-sep 5262, ax-nul 5274, ax-pr 5409 and shorten proof. (Revised by BJ, 14-Apr-2024.)

Assertion

Ref	Expression
pwundif	⊢ 𝒫 (𝐴 ∪ 𝐵) = ((𝒫 (𝐴 ∪ 𝐵) ∖ 𝒫 𝐴) ∪ 𝒫 𝐴)

Proof of Theorem pwundif

Step	Hyp	Ref	Expression
1		ssun1 4134	. . . 4 ⊢ 𝐴 ⊆ (𝐴 ∪ 𝐵)
2	1	sspwi 4579	. . 3 ⊢ 𝒫 𝐴 ⊆ 𝒫 (𝐴 ∪ 𝐵)
3		undif 4448	. . 3 ⊢ (𝒫 𝐴 ⊆ 𝒫 (𝐴 ∪ 𝐵) ↔ (𝒫 𝐴 ∪ (𝒫 (𝐴 ∪ 𝐵) ∖ 𝒫 𝐴)) = 𝒫 (𝐴 ∪ 𝐵))
4	2, 3	mpbi 233	. 2 ⊢ (𝒫 𝐴 ∪ (𝒫 (𝐴 ∪ 𝐵) ∖ 𝒫 𝐴)) = 𝒫 (𝐴 ∪ 𝐵)
5		uncom 4115	. 2 ⊢ (𝒫 𝐴 ∪ (𝒫 (𝐴 ∪ 𝐵) ∖ 𝒫 𝐴)) = ((𝒫 (𝐴 ∪ 𝐵) ∖ 𝒫 𝐴) ∪ 𝒫 𝐴)
6	4, 5	eqtr3i 2791	1 ⊢ 𝒫 (𝐴 ∪ 𝐵) = ((𝒫 (𝐴 ∪ 𝐵) ∖ 𝒫 𝐴) ∪ 𝒫 𝐴)

This proof depends on syntax axioms:   = wceq 1570   ∖ cdif 3905   ∪ cun 3906   ⊆ wss 3908  𝒫 cpw 4567

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 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738

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 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-rab 3420  df-v 3460  df-dif 3911  df-un 3913  df-in 3915  df-ss 3925  df-nul 4290  df-pw 4569

This theorem is used by:  pwfilem  9287