Theorem inindi 4131

Description: Intersection distributes over itself. (Contributed by NM, 6-May-1994.)

Assertion

Ref	Expression
inindi	⊢ (𝐴 ∩ (𝐵 ∩ 𝐶)) = ((𝐴 ∩ 𝐵) ∩ (𝐴 ∩ 𝐶))

Proof of Theorem inindi

Step	Hyp	Ref	Expression
1		inidm 4123	. . 3 ⊢ (𝐴 ∩ 𝐴) = 𝐴
2	1	ineq1i 4113	. 2 ⊢ ((𝐴 ∩ 𝐴) ∩ (𝐵 ∩ 𝐶)) = (𝐴 ∩ (𝐵 ∩ 𝐶))
3		in4 4130	. 2 ⊢ ((𝐴 ∩ 𝐴) ∩ (𝐵 ∩ 𝐶)) = ((𝐴 ∩ 𝐵) ∩ (𝐴 ∩ 𝐶))
4	2, 3	eqtr3i 2785	1 ⊢ (𝐴 ∩ (𝐵 ∩ 𝐶)) = ((𝐴 ∩ 𝐵) ∩ (𝐴 ∩ 𝐶))

This proof depends on syntax axioms:   = wceq 1569   ∩ cin 4089

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-in 4091

This theorem is used by:  difundi  4235  dfif5  4498  resindi  5985  offres  7995  incexclem  16005  bitsinv1  16612  bitsinvp1  16619  bitsres  16643  fh1  32220