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Theorem indif1 3209
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 3208 . 2 (𝐵 ∩ (𝐴𝐶)) = ((𝐵𝐴) ∖ 𝐶)
2 incom 3158 . 2 (𝐵 ∩ (𝐴𝐶)) = ((𝐴𝐶) ∩ 𝐵)
3 incom 3158 . . 3 (𝐵𝐴) = (𝐴𝐵)
43difeq1i 3086 . 2 ((𝐵𝐴) ∖ 𝐶) = ((𝐴𝐵) ∖ 𝐶)
51, 2, 43eqtr3i 2109 1 ((𝐴𝐶) ∩ 𝐵) = ((𝐴𝐵) ∖ 𝐶)
Colors of variables: wff set class
Syntax hints:   = wceq 1284  cdif 2970  cin 2972
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 576  ax-in2 577  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063
This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-rab 2357  df-v 2603  df-dif 2975  df-in 2979
This theorem is referenced by:  resdmdfsn  4671
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