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Theorem indif1 4160
Description: Bring an intersection in and out of a class difference. (Contributed by Mario Carneiro, 15-May-2015.)
Assertion
Ref Expression
indif1 ((𝐴𝐶) ∩ 𝐵) = ((𝐴𝐵) ∖ 𝐶)

Proof of Theorem indif1
StepHypRef Expression
1 indif2 4159 . 2 (𝐵 ∩ (𝐴𝐶)) = ((𝐵𝐴) ∖ 𝐶)
2 incom 4089 . 2 (𝐵 ∩ (𝐴𝐶)) = ((𝐴𝐶) ∩ 𝐵)
3 incom 4089 . . 3 (𝐵𝐴) = (𝐴𝐵)
43difeq1i 4007 . 2 ((𝐵𝐴) ∖ 𝐶) = ((𝐴𝐵) ∖ 𝐶)
51, 2, 43eqtr3i 2769 1 ((𝐴𝐶) ∩ 𝐵) = ((𝐴𝐵) ∖ 𝐶)
Colors of variables: wff setvar class
Syntax hints:   = wceq 1542  cdif 3838  cin 3840
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1916  ax-6 1974  ax-7 2019  ax-8 2115  ax-9 2123  ax-ext 2710
This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1545  df-ex 1787  df-sb 2074  df-clab 2717  df-cleq 2730  df-clel 2811  df-rab 3062  df-v 3399  df-dif 3844  df-in 3848
This theorem is referenced by:  resdifcom  5838  resdmdfsn  5869  hartogslem1  9072  fpwwe2  10136  leiso  13904  basdif0  21697  tgdif0  21736  kqdisj  22476  trufil  22654  difininv  30430  gtiso  30600  dfon4  33825  disjdifb  45671
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