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Theorem indif1 4250
Description: Bring an intersection in and out of a class difference. (Contributed by Mario Carneiro, 15-May-2015.)
Assertion
Ref Expression
indif1 ((𝐴𝐶) ∩ 𝐵) = ((𝐴𝐵) ∖ 𝐶)

Proof of Theorem indif1
StepHypRef Expression
1 indif2 4249 . 2 (𝐵 ∩ (𝐴𝐶)) = ((𝐵𝐴) ∖ 𝐶)
2 incom 4180 . 2 (𝐵 ∩ (𝐴𝐶)) = ((𝐴𝐶) ∩ 𝐵)
3 incom 4180 . . 3 (𝐵𝐴) = (𝐴𝐵)
43difeq1i 4097 . 2 ((𝐵𝐴) ∖ 𝐶) = ((𝐴𝐵) ∖ 𝐶)
51, 2, 43eqtr3i 2854 1 ((𝐴𝐶) ∩ 𝐵) = ((𝐴𝐵) ∖ 𝐶)
Colors of variables: wff setvar class
Syntax hints:   = wceq 1537  cdif 3935  cin 3937
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2795
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2802  df-cleq 2816  df-clel 2895  df-nfc 2965  df-rab 3149  df-v 3498  df-dif 3941  df-in 3945
This theorem is referenced by:  resdifcom  5874  resdmdfsn  5903  hartogslem1  9008  fpwwe2  10067  leiso  13820  basdif0  21563  tgdif0  21602  kqdisj  22342  trufil  22520  difininv  30281  gtiso  30438  dfon4  33356
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