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Theorem indif2 3290
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 3256 . 2 ((𝐴𝐵) ∩ (V ∖ 𝐶)) = (𝐴 ∩ (𝐵 ∩ (V ∖ 𝐶)))
2 invdif 3288 . 2 ((𝐴𝐵) ∩ (V ∖ 𝐶)) = ((𝐴𝐵) ∖ 𝐶)
3 invdif 3288 . . 3 (𝐵 ∩ (V ∖ 𝐶)) = (𝐵𝐶)
43ineq2i 3244 . 2 (𝐴 ∩ (𝐵 ∩ (V ∖ 𝐶))) = (𝐴 ∩ (𝐵𝐶))
51, 2, 43eqtr3ri 2147 1 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐵) ∖ 𝐶)
Colors of variables: wff set class
Syntax hints:   = wceq 1316  Vcvv 2660  cdif 3038  cin 3040
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 588  ax-in2 589  ax-io 683  ax-5 1408  ax-7 1409  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-8 1467  ax-10 1468  ax-11 1469  ax-i12 1470  ax-bndl 1471  ax-4 1472  ax-17 1491  ax-i9 1495  ax-ial 1499  ax-i5r 1500  ax-ext 2099
This theorem depends on definitions:  df-bi 116  df-tru 1319  df-nf 1422  df-sb 1721  df-clab 2104  df-cleq 2110  df-clel 2113  df-nfc 2247  df-v 2662  df-dif 3043  df-in 3047
This theorem is referenced by:  indif1  3291  indifcom  3292  difopn  12188
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