Theorem in4 4159

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 4154	. . 3 ⊢ (𝐵 ∩ (𝐶 ∩ 𝐷)) = (𝐶 ∩ (𝐵 ∩ 𝐷))
2	1	ineq2i 4143	. 2 ⊢ (𝐴 ∩ (𝐵 ∩ (𝐶 ∩ 𝐷))) = (𝐴 ∩ (𝐶 ∩ (𝐵 ∩ 𝐷)))
3		inass 4153	. 2 ⊢ ((𝐴 ∩ 𝐵) ∩ (𝐶 ∩ 𝐷)) = (𝐴 ∩ (𝐵 ∩ (𝐶 ∩ 𝐷)))
4		inass 4153	. 2 ⊢ ((𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐷)) = (𝐴 ∩ (𝐶 ∩ (𝐵 ∩ 𝐷)))
5	2, 3, 4	3eqtr4i 2776	1 ⊢ ((𝐴 ∩ 𝐵) ∩ (𝐶 ∩ 𝐷)) = ((𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐷))

Syntax hints:   = wceq 1539   ∩ cin 3886

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-rab 3073  df-v 3434  df-in 3894

This theorem is referenced by:  inindi  4160  inindir  4161  fh2  29981  disjxpin  30927