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Theorem indif2 3288
Description: Bring an intersection in and out of a class difference. (Contributed by Jeff Hankins, 15-Jul-2009.)
Assertion
Ref Expression
indif2 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐵) ∖ 𝐶)

Proof of Theorem indif2
StepHypRef Expression
1 inass 3254 . 2 ((𝐴𝐵) ∩ (V ∖ 𝐶)) = (𝐴 ∩ (𝐵 ∩ (V ∖ 𝐶)))
2 invdif 3286 . 2 ((𝐴𝐵) ∩ (V ∖ 𝐶)) = ((𝐴𝐵) ∖ 𝐶)
3 invdif 3286 . . 3 (𝐵 ∩ (V ∖ 𝐶)) = (𝐵𝐶)
43ineq2i 3242 . 2 (𝐴 ∩ (𝐵 ∩ (V ∖ 𝐶))) = (𝐴 ∩ (𝐵𝐶))
51, 2, 43eqtr3ri 2145 1 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐵) ∖ 𝐶)
Colors of variables: wff set class
Syntax hints:   = wceq 1314  Vcvv 2658  cdif 3036  cin 3038
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 586  ax-in2 587  ax-io 681  ax-5 1406  ax-7 1407  ax-gen 1408  ax-ie1 1452  ax-ie2 1453  ax-8 1465  ax-10 1466  ax-11 1467  ax-i12 1468  ax-bndl 1469  ax-4 1470  ax-17 1489  ax-i9 1493  ax-ial 1497  ax-i5r 1498  ax-ext 2097
This theorem depends on definitions:  df-bi 116  df-tru 1317  df-nf 1420  df-sb 1719  df-clab 2102  df-cleq 2108  df-clel 2111  df-nfc 2245  df-v 2660  df-dif 3041  df-in 3045
This theorem is referenced by:  indif1  3289  indifcom  3290  difopn  12183
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