Theorem indifdir 4301

Description: Distribute intersection over difference. (Contributed by Scott Fenton, 14-Apr-2011.) (Revised by BTernaryTau, 14-Aug-2024.)

Assertion

Ref	Expression
indifdir	⊢ ((𝐴 ∖ 𝐵) ∩ 𝐶) = ((𝐴 ∩ 𝐶) ∖ (𝐵 ∩ 𝐶))

Proof of Theorem indifdir

Step	Hyp	Ref	Expression
1		indifdi 4300	. 2 ⊢ (𝐶 ∩ (𝐴 ∖ 𝐵)) = ((𝐶 ∩ 𝐴) ∖ (𝐶 ∩ 𝐵))
2		incom 4217	. 2 ⊢ ((𝐴 ∖ 𝐵) ∩ 𝐶) = (𝐶 ∩ (𝐴 ∖ 𝐵))
3		incom 4217	. . 3 ⊢ (𝐴 ∩ 𝐶) = (𝐶 ∩ 𝐴)
4		incom 4217	. . 3 ⊢ (𝐵 ∩ 𝐶) = (𝐶 ∩ 𝐵)
5	3, 4	difeq12i 4134	. 2 ⊢ ((𝐴 ∩ 𝐶) ∖ (𝐵 ∩ 𝐶)) = ((𝐶 ∩ 𝐴) ∖ (𝐶 ∩ 𝐵))
6	1, 2, 5	3eqtr4i 2773	1 ⊢ ((𝐴 ∖ 𝐵) ∩ 𝐶) = ((𝐴 ∩ 𝐶) ∖ (𝐵 ∩ 𝐶))

Syntax hints:   = wceq 1537   ∖ cdif 3960   ∩ cin 3962

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-rab 3434  df-v 3480  df-dif 3966  df-in 3970

This theorem is referenced by:  resdifdir  6259  preddif  6352  fresaun  6780  uniioombllem4  25635  subsalsal  46315