Theorem pwundif 4582

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 5248, ax-nul 5259, ax-pr 5390 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 4176	. . . 4 ⊢ 𝐴 ⊆ (𝐴 ∪ 𝐵)
2	1	sspwi 4568	. . 3 ⊢ 𝒫 𝐴 ⊆ 𝒫 (𝐴 ∪ 𝐵)
3		undif 4437	. . 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 4157	. 2 ⊢ (𝒫 𝐴 ∪ (𝒫 (𝐴 ∪ 𝐵) ∖ 𝒫 𝐴)) = ((𝒫 (𝐴 ∪ 𝐵) ∖ 𝒫 𝐴) ∪ 𝒫 𝐴)
6	4, 5	eqtr3i 2785	1 ⊢ 𝒫 (𝐴 ∪ 𝐵) = ((𝒫 (𝐴 ∪ 𝐵) ∖ 𝒫 𝐴) ∪ 𝒫 𝐴)

This proof depends on syntax axioms:   = wceq 1569   ⊆ wss 3895   ∖ cdif 4045   ∪ cun 4149  𝒫 cpw 4556

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-ss 3896  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279  df-pw 4558

This theorem is used by:  pwfilem  9312