Theorem pwundif 4588

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 5258, ax-nul 5270, ax-pr 5406 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 4132	. . . 4 ⊢ 𝐴 ⊆ (𝐴 ∪ 𝐵)
2	1	sspwi 4575	. . 3 ⊢ 𝒫 𝐴 ⊆ 𝒫 (𝐴 ∪ 𝐵)
3		undif 4444	. . 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 4113	. 2 ⊢ (𝒫 𝐴 ∪ (𝒫 (𝐴 ∪ 𝐵) ∖ 𝒫 𝐴)) = ((𝒫 (𝐴 ∪ 𝐵) ∖ 𝒫 𝐴) ∪ 𝒫 𝐴)
6	4, 5	eqtr3i 2788	1 ⊢ 𝒫 (𝐴 ∪ 𝐵) = ((𝒫 (𝐴 ∪ 𝐵) ∖ 𝒫 𝐴) ∪ 𝒫 𝐴)

Syntax hints:   = wceq 1570   ∖ cdif 3903   ∪ cun 3904   ⊆ wss 3906  𝒫 cpw 4563

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-rab 3417  df-v 3457  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4288  df-pw 4565

This theorem is referenced by:  pwfilem  9278