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Theorem in31 3198
Description: A rearrangement of intersection. (Contributed by NM, 27-Aug-2012.)
Assertion
Ref Expression
in31 ((𝐴𝐵) ∩ 𝐶) = ((𝐶𝐵) ∩ 𝐴)

Proof of Theorem in31
StepHypRef Expression
1 in12 3195 . 2 (𝐶 ∩ (𝐴𝐵)) = (𝐴 ∩ (𝐶𝐵))
2 incom 3176 . 2 ((𝐴𝐵) ∩ 𝐶) = (𝐶 ∩ (𝐴𝐵))
3 incom 3176 . 2 ((𝐶𝐵) ∩ 𝐴) = (𝐴 ∩ (𝐶𝐵))
41, 2, 33eqtr4i 2113 1 ((𝐴𝐵) ∩ 𝐶) = ((𝐶𝐵) ∩ 𝐴)
Colors of variables: wff set class
Syntax hints:   = wceq 1285  cin 2983
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-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2065
This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1688  df-clab 2070  df-cleq 2076  df-clel 2079  df-nfc 2212  df-v 2614  df-in 2990
This theorem is referenced by:  inrot  3199
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