Theorem in4 4130

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 4125	. . 3 ⊢ (𝐵 ∩ (𝐶 ∩ 𝐷)) = (𝐶 ∩ (𝐵 ∩ 𝐷))
2	1	ineq2i 4114	. 2 ⊢ (𝐴 ∩ (𝐵 ∩ (𝐶 ∩ 𝐷))) = (𝐴 ∩ (𝐶 ∩ (𝐵 ∩ 𝐷)))
3		inass 4124	. 2 ⊢ ((𝐴 ∩ 𝐵) ∩ (𝐶 ∩ 𝐷)) = (𝐴 ∩ (𝐵 ∩ (𝐶 ∩ 𝐷)))
4		inass 4124	. 2 ⊢ ((𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐷)) = (𝐴 ∩ (𝐶 ∩ (𝐵 ∩ 𝐷)))
5	2, 3, 4	3eqtr4i 2793	1 ⊢ ((𝐴 ∩ 𝐵) ∩ (𝐶 ∩ 𝐷)) = ((𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐷))

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:  inindi  4131  inindir  4132  fh2  32221  disjxpin  33224