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Theorem indif1 3499
 Description: Bring an intersection in and out of a class difference. (Contributed by Mario Carneiro, 15-May-2015.)
Assertion
Ref Expression
indif1 ((A C) ∩ B) = ((AB) C)

Proof of Theorem indif1
StepHypRef Expression
1 indif2 3498 . 2 (B ∩ (A C)) = ((BA) C)
2 incom 3448 . 2 (B ∩ (A C)) = ((A C) ∩ B)
3 incom 3448 . . 3 (BA) = (AB)
43difeq1i 3381 . 2 ((BA) C) = ((AB) C)
51, 2, 43eqtr3i 2381 1 ((A C) ∩ B) = ((AB) C)
 Colors of variables: wff setvar class Syntax hints:   = wceq 1642   ∖ cdif 3206   ∩ cin 3208 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-dif 3215 This theorem is referenced by: (None)
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