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Theorem indifundif 29200
Description: A remarkable equation with sets. (Contributed by Thierry Arnoux, 18-May-2020.)
Assertion
Ref Expression
indifundif (((𝐴𝐵) ∖ 𝐶) ∪ (𝐴𝐵)) = (𝐴 ∖ (𝐵𝐶))

Proof of Theorem indifundif
StepHypRef Expression
1 difindi 3857 . 2 (𝐴 ∖ (𝐵𝐶)) = ((𝐴𝐵) ∪ (𝐴𝐶))
2 difundir 3856 . . . . 5 (((𝐴𝐵) ∪ (𝐴𝐵)) ∖ 𝐶) = (((𝐴𝐵) ∖ 𝐶) ∪ ((𝐴𝐵) ∖ 𝐶))
3 inundif 4018 . . . . . 6 ((𝐴𝐵) ∪ (𝐴𝐵)) = 𝐴
43difeq1i 3702 . . . . 5 (((𝐴𝐵) ∪ (𝐴𝐵)) ∖ 𝐶) = (𝐴𝐶)
5 uncom 3735 . . . . 5 (((𝐴𝐵) ∖ 𝐶) ∪ ((𝐴𝐵) ∖ 𝐶)) = (((𝐴𝐵) ∖ 𝐶) ∪ ((𝐴𝐵) ∖ 𝐶))
62, 4, 53eqtr3i 2651 . . . 4 (𝐴𝐶) = (((𝐴𝐵) ∖ 𝐶) ∪ ((𝐴𝐵) ∖ 𝐶))
76uneq2i 3742 . . 3 ((𝐴𝐵) ∪ (𝐴𝐶)) = ((𝐴𝐵) ∪ (((𝐴𝐵) ∖ 𝐶) ∪ ((𝐴𝐵) ∖ 𝐶)))
8 unass 3748 . . 3 (((𝐴𝐵) ∪ ((𝐴𝐵) ∖ 𝐶)) ∪ ((𝐴𝐵) ∖ 𝐶)) = ((𝐴𝐵) ∪ (((𝐴𝐵) ∖ 𝐶) ∪ ((𝐴𝐵) ∖ 𝐶)))
9 undifabs 4017 . . . 4 ((𝐴𝐵) ∪ ((𝐴𝐵) ∖ 𝐶)) = (𝐴𝐵)
109uneq1i 3741 . . 3 (((𝐴𝐵) ∪ ((𝐴𝐵) ∖ 𝐶)) ∪ ((𝐴𝐵) ∖ 𝐶)) = ((𝐴𝐵) ∪ ((𝐴𝐵) ∖ 𝐶))
117, 8, 103eqtr2i 2649 . 2 ((𝐴𝐵) ∪ (𝐴𝐶)) = ((𝐴𝐵) ∪ ((𝐴𝐵) ∖ 𝐶))
12 uncom 3735 . 2 ((𝐴𝐵) ∪ ((𝐴𝐵) ∖ 𝐶)) = (((𝐴𝐵) ∖ 𝐶) ∪ (𝐴𝐵))
131, 11, 123eqtrri 2648 1 (((𝐴𝐵) ∖ 𝐶) ∪ (𝐴𝐵)) = (𝐴 ∖ (𝐵𝐶))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1480  cdif 3552  cun 3553  cin 3554
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2750  df-ral 2912  df-rab 2916  df-v 3188  df-dif 3558  df-un 3560  df-in 3562
This theorem is referenced by:  inelcarsg  30151
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