Theorem indifundif 33120

Description: A remarkable equation with sets. (Contributed by Thierry Arnoux, 18-May-2020.)

Assertion

Ref	Expression
indifundif	⊢ (((𝐴 ∩ 𝐵) ∖ 𝐶) ∪ (𝐴 ∖ 𝐵)) = (𝐴 ∖ (𝐵 ∩ 𝐶))

Proof of Theorem indifundif

Step	Hyp	Ref	Expression
1		difindi 4237	. 2 ⊢ (𝐴 ∖ (𝐵 ∩ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ (𝐴 ∖ 𝐶))
2		difundir 4236	. . . . 5 ⊢ (((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) ∖ 𝐶) = (((𝐴 ∩ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶))
3		inundif 4434	. . . . . 6 ⊢ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) = 𝐴
4	3	difeq1i 4059	. . . . 5 ⊢ (((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) ∖ 𝐶) = (𝐴 ∖ 𝐶)
5		uncom 4157	. . . . 5 ⊢ (((𝐴 ∩ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶)) = (((𝐴 ∖ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶))
6	2, 4, 5	3eqtr3i 2791	. . . 4 ⊢ (𝐴 ∖ 𝐶) = (((𝐴 ∖ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶))
7	6	uneq2i 4164	. . 3 ⊢ ((𝐴 ∖ 𝐵) ∪ (𝐴 ∖ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ (((𝐴 ∖ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)))
8		unass 4170	. . 3 ⊢ (((𝐴 ∖ 𝐵) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶)) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ (((𝐴 ∖ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)))
9		undifabs 4433	. . . 4 ⊢ ((𝐴 ∖ 𝐵) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶)) = (𝐴 ∖ 𝐵)
10	9	uneq1i 4163	. . 3 ⊢ (((𝐴 ∖ 𝐵) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶)) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶))
11	7, 8, 10	3eqtr2i 2789	. 2 ⊢ ((𝐴 ∖ 𝐵) ∪ (𝐴 ∖ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶))
12		uncom 4157	. 2 ⊢ ((𝐴 ∖ 𝐵) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)) = (((𝐴 ∩ 𝐵) ∖ 𝐶) ∪ (𝐴 ∖ 𝐵))
13	1, 11, 12	3eqtrri 2788	1 ⊢ (((𝐴 ∩ 𝐵) ∖ 𝐶) ∪ (𝐴 ∖ 𝐵)) = (𝐴 ∖ (𝐵 ∩ 𝐶))

This proof depends on syntax axioms:   = wceq 1569   ∖ cdif 4045   ∩ cin 4089   ∪ cun 4149

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-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-dif 4047  df-in 4091  df-un 4151

This theorem is used by:  inelcarsg  34943