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Theorem indif2 3371
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 3337 . 2 ((𝐴𝐵) ∩ (V ∖ 𝐶)) = (𝐴 ∩ (𝐵 ∩ (V ∖ 𝐶)))
2 invdif 3369 . 2 ((𝐴𝐵) ∩ (V ∖ 𝐶)) = ((𝐴𝐵) ∖ 𝐶)
3 invdif 3369 . . 3 (𝐵 ∩ (V ∖ 𝐶)) = (𝐵𝐶)
43ineq2i 3325 . 2 (𝐴 ∩ (𝐵 ∩ (V ∖ 𝐶))) = (𝐴 ∩ (𝐵𝐶))
51, 2, 43eqtr3ri 2200 1 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐵) ∖ 𝐶)
Colors of variables: wff set class
Syntax hints:   = wceq 1348  Vcvv 2730  cdif 3118  cin 3120
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 609  ax-in2 610  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152
This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-dif 3123  df-in 3127
This theorem is referenced by:  indif1  3372  indifcom  3373  difopn  12863
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