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Theorem indif2 3425
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 3391 . 2 ((𝐴𝐵) ∩ (V ∖ 𝐶)) = (𝐴 ∩ (𝐵 ∩ (V ∖ 𝐶)))
2 invdif 3423 . 2 ((𝐴𝐵) ∩ (V ∖ 𝐶)) = ((𝐴𝐵) ∖ 𝐶)
3 invdif 3423 . . 3 (𝐵 ∩ (V ∖ 𝐶)) = (𝐵𝐶)
43ineq2i 3379 . 2 (𝐴 ∩ (𝐵 ∩ (V ∖ 𝐶))) = (𝐴 ∩ (𝐵𝐶))
51, 2, 43eqtr3ri 2237 1 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐵) ∖ 𝐶)
Colors of variables: wff set class
Syntax hints:   = wceq 1373  Vcvv 2776  cdif 3171  cin 3173
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2189
This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2194  df-cleq 2200  df-clel 2203  df-nfc 2339  df-v 2778  df-dif 3176  df-in 3180
This theorem is referenced by:  indif1  3426  indifcom  3427  difopn  14695
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