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Theorem indif1 4101
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 4100 . 2 (𝐵 ∩ (𝐴𝐶)) = ((𝐵𝐴) ∖ 𝐶)
2 incom 4032 . 2 (𝐵 ∩ (𝐴𝐶)) = ((𝐴𝐶) ∩ 𝐵)
3 incom 4032 . . 3 (𝐵𝐴) = (𝐴𝐵)
43difeq1i 3951 . 2 ((𝐵𝐴) ∖ 𝐶) = ((𝐴𝐵) ∖ 𝐶)
51, 2, 43eqtr3i 2857 1 ((𝐴𝐶) ∩ 𝐵) = ((𝐴𝐵) ∖ 𝐶)
Colors of variables: wff setvar class
Syntax hints:   = wceq 1656  cdif 3795  cin 3797
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1894  ax-4 1908  ax-5 2009  ax-6 2075  ax-7 2112  ax-9 2173  ax-10 2192  ax-11 2207  ax-12 2220  ax-ext 2803
This theorem depends on definitions:  df-bi 199  df-an 387  df-or 879  df-tru 1660  df-ex 1879  df-nf 1883  df-sb 2068  df-clab 2812  df-cleq 2818  df-clel 2821  df-nfc 2958  df-ral 3122  df-rab 3126  df-v 3416  df-dif 3801  df-in 3805
This theorem is referenced by:  resdifcom  5652  resdmdfsn  5682  hartogslem1  8716  fpwwe2  9780  leiso  13532  basdif0  21128  tgdif0  21167  kqdisj  21906  trufil  22084  difininv  29891  gtiso  30015  dfon4  32528
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