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Theorem indif2 3475
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 3441 . 2 ((𝐴𝐵) ∩ (V ∖ 𝐶)) = (𝐴 ∩ (𝐵 ∩ (V ∖ 𝐶)))
2 invdif 3473 . 2 ((𝐴𝐵) ∩ (V ∖ 𝐶)) = ((𝐴𝐵) ∖ 𝐶)
3 invdif 3473 . . 3 (𝐵 ∩ (V ∖ 𝐶)) = (𝐵𝐶)
43ineq2i 3429 . 2 (𝐴 ∩ (𝐵 ∩ (V ∖ 𝐶))) = (𝐴 ∩ (𝐵𝐶))
51, 2, 43eqtr3ri 2268 1 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐵) ∖ 𝐶)
Colors of variables: wff set class
Syntax hints:   = wceq 1402  Vcvv 2821  cdif 3217  cin 3219
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 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220
This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-dif 3222  df-in 3226
This theorem is referenced by:  indif1  3476  indifcom  3477  inssdif0im  3592  difopn  15192
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