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Theorem inindir 4199
Description: Intersection distributes over itself. (Contributed by NM, 17-Aug-2004.)
Assertion
Ref Expression
inindir ((𝐴𝐵) ∩ 𝐶) = ((𝐴𝐶) ∩ (𝐵𝐶))

Proof of Theorem inindir
StepHypRef Expression
1 inidm 4190 . . 3 (𝐶𝐶) = 𝐶
21ineq2i 4180 . 2 ((𝐴𝐵) ∩ (𝐶𝐶)) = ((𝐴𝐵) ∩ 𝐶)
3 in4 4197 . 2 ((𝐴𝐵) ∩ (𝐶𝐶)) = ((𝐴𝐶) ∩ (𝐵𝐶))
42, 3eqtr3i 2754 1 ((𝐴𝐵) ∩ 𝐶) = ((𝐴𝐶) ∩ (𝐵𝐶))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1540  cin 3913
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2701
This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-rab 3406  df-v 3449  df-in 3921
This theorem is referenced by:  difindir  4256  resindir  5967  predin  6300  restbas  23045  connsuba  23307  kgentopon  23425  trfbas2  23730  trfil2  23774  fclsrest  23911  trust  24117  chtdif  27068  ppidif  27073  mdslmd1lem1  32254  mdslmd1lem2  32255  mddmdin0i  32360  ballotlemgun  34516  cvmsss2  35261
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