Theorem indifundif 32881

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 4244	. 2 ⊢ (𝐴 ∖ (𝐵 ∩ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ (𝐴 ∖ 𝐶))
2		difundir 4243	. . . . 5 ⊢ (((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) ∖ 𝐶) = (((𝐴 ∩ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶))
3		inundif 4439	. . . . . 6 ⊢ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) = 𝐴
4	3	difeq1i 4076	. . . . 5 ⊢ (((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) ∖ 𝐶) = (𝐴 ∖ 𝐶)
5		uncom 4111	. . . . 5 ⊢ (((𝐴 ∩ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶)) = (((𝐴 ∖ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶))
6	2, 4, 5	3eqtr3i 2793	. . . 4 ⊢ (𝐴 ∖ 𝐶) = (((𝐴 ∖ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶))
7	6	uneq2i 4118	. . 3 ⊢ ((𝐴 ∖ 𝐵) ∪ (𝐴 ∖ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ (((𝐴 ∖ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)))
8		unass 4124	. . 3 ⊢ (((𝐴 ∖ 𝐵) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶)) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ (((𝐴 ∖ 𝐵) ∖ 𝐶) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)))
9		undifabs 4438	. . . 4 ⊢ ((𝐴 ∖ 𝐵) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶)) = (𝐴 ∖ 𝐵)
10	9	uneq1i 4117	. . 3 ⊢ (((𝐴 ∖ 𝐵) ∪ ((𝐴 ∖ 𝐵) ∖ 𝐶)) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶))
11	7, 8, 10	3eqtr2i 2791	. 2 ⊢ ((𝐴 ∖ 𝐵) ∪ (𝐴 ∖ 𝐶)) = ((𝐴 ∖ 𝐵) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶))
12		uncom 4111	. 2 ⊢ ((𝐴 ∖ 𝐵) ∪ ((𝐴 ∩ 𝐵) ∖ 𝐶)) = (((𝐴 ∩ 𝐵) ∖ 𝐶) ∪ (𝐴 ∖ 𝐵))
13	1, 11, 12	3eqtrri 2790	1 ⊢ (((𝐴 ∩ 𝐵) ∖ 𝐶) ∪ (𝐴 ∖ 𝐵)) = (𝐴 ∖ (𝐵 ∩ 𝐶))

This proof depends on syntax axioms:   = wceq 1569   ∖ cdif 3901   ∪ cun 3902   ∩ cin 3903

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3416  df-v 3456  df-dif 3907  df-un 3909  df-in 3911

This theorem is used by:  inelcarsg  34710