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Theorem indif1 3182
 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 3181 . 2 (𝐵 ∩ (𝐴𝐶)) = ((𝐵𝐴) ∖ 𝐶)
2 incom 3129 . 2 (𝐵 ∩ (𝐴𝐶)) = ((𝐴𝐶) ∩ 𝐵)
3 incom 3129 . . 3 (𝐵𝐴) = (𝐴𝐵)
43difeq1i 3058 . 2 ((𝐵𝐴) ∖ 𝐶) = ((𝐴𝐵) ∖ 𝐶)
51, 2, 43eqtr3i 2068 1 ((𝐴𝐶) ∩ 𝐵) = ((𝐴𝐵) ∖ 𝐶)
 Colors of variables: wff set class Syntax hints:   = wceq 1243   ∖ cdif 2914   ∩ cin 2916 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022 This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-rab 2315  df-v 2559  df-dif 2920  df-in 2924 This theorem is referenced by: (None)
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