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Theorem indif1 3528
Description: Bring an intersection in and out of a class difference. (Contributed by Mario Carneiro, 15-May-2015.)
Assertion
Ref Expression
indif1  |-  ( ( A  \  C )  i^i  B )  =  ( ( A  i^i  B )  \  C )

Proof of Theorem indif1
StepHypRef Expression
1 indif2 3527 . 2  |-  ( B  i^i  ( A  \  C ) )  =  ( ( B  i^i  A )  \  C )
2 incom 3476 . 2  |-  ( B  i^i  ( A  \  C ) )  =  ( ( A  \  C )  i^i  B
)
3 incom 3476 . . 3  |-  ( B  i^i  A )  =  ( A  i^i  B
)
43difeq1i 3404 . 2  |-  ( ( B  i^i  A ) 
\  C )  =  ( ( A  i^i  B )  \  C )
51, 2, 43eqtr3i 2415 1  |-  ( ( A  \  C )  i^i  B )  =  ( ( A  i^i  B )  \  C )
Colors of variables: wff set class
Syntax hints:    = wceq 1649    \ cdif 3260    i^i cin 3262
This theorem is referenced by:  hartogslem1  7444  fpwwe2  8451  leiso  11635  basdif0  16941  tgdif0  16980  kqdisj  17685  trufil  17863  gtiso  23929  dfon4  25457
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1552  ax-5 1563  ax-17 1623  ax-9 1661  ax-8 1682  ax-6 1736  ax-7 1741  ax-11 1753  ax-12 1939  ax-ext 2368
This theorem depends on definitions:  df-bi 178  df-or 360  df-an 361  df-tru 1325  df-ex 1548  df-nf 1551  df-sb 1656  df-clab 2374  df-cleq 2380  df-clel 2383  df-nfc 2512  df-ral 2654  df-rab 2658  df-v 2901  df-dif 3266  df-in 3270
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