Theorem in4 4200

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 4195	. . 3 ⊢ (𝐵 ∩ (𝐶 ∩ 𝐷)) = (𝐶 ∩ (𝐵 ∩ 𝐷))
2	1	ineq2i 4183	. 2 ⊢ (𝐴 ∩ (𝐵 ∩ (𝐶 ∩ 𝐷))) = (𝐴 ∩ (𝐶 ∩ (𝐵 ∩ 𝐷)))
3		inass 4194	. 2 ⊢ ((𝐴 ∩ 𝐵) ∩ (𝐶 ∩ 𝐷)) = (𝐴 ∩ (𝐵 ∩ (𝐶 ∩ 𝐷)))
4		inass 4194	. 2 ⊢ ((𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐷)) = (𝐴 ∩ (𝐶 ∩ (𝐵 ∩ 𝐷)))
5	2, 3, 4	3eqtr4i 2763	1 ⊢ ((𝐴 ∩ 𝐵) ∩ (𝐶 ∩ 𝐷)) = ((𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐷))

Syntax hints:   = wceq 1540   ∩ cin 3916

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-rab 3409  df-v 3452  df-in 3924

This theorem is referenced by:  inindi  4201  inindir  4202  fh2  31555  disjxpin  32524