Theorem in4 4185

Description: Rearrangement of intersection of 4 classes. (Contributed by NM, 21-Apr-2001.)

Assertion

Ref	Expression
in4	⊢ ((𝐴 ∩ 𝐵) ∩ (𝐶 ∩ 𝐷)) = ((𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐷))

Proof of Theorem in4

Step	Hyp	Ref	Expression
1		in12 4180	. . 3 ⊢ (𝐵 ∩ (𝐶 ∩ 𝐷)) = (𝐶 ∩ (𝐵 ∩ 𝐷))
2	1	ineq2i 4169	. 2 ⊢ (𝐴 ∩ (𝐵 ∩ (𝐶 ∩ 𝐷))) = (𝐴 ∩ (𝐶 ∩ (𝐵 ∩ 𝐷)))
3		inass 4179	. 2 ⊢ ((𝐴 ∩ 𝐵) ∩ (𝐶 ∩ 𝐷)) = (𝐴 ∩ (𝐵 ∩ (𝐶 ∩ 𝐷)))
4		inass 4179	. 2 ⊢ ((𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐷)) = (𝐴 ∩ (𝐶 ∩ (𝐵 ∩ 𝐷)))
5	2, 3, 4	3eqtr4i 2795	1 ⊢ ((𝐴 ∩ 𝐵) ∩ (𝐶 ∩ 𝐷)) = ((𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐷))

This proof depends on syntax axioms:   = wceq 1569   ∩ cin 3903

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3416  df-v 3456  df-in 3911

This theorem is used by:  inindi  4186  inindir  4187  fh2  31982  disjxpin  32944