Theorem inindir 1655

Description: Intersection distributes over itself.

Assertion

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

Proof of Theorem inindir

Step	Hyp	Ref	Expression
1		inidm 1649	. . 3 ⊢ (C ∩ C) = C
2	1	ineq2i 1642	. 2 ⊢ ((A ∩ B) ∩ (C ∩ C)) = ((A ∩ B) ∩ C)
3		in4 1653	. 2 ⊢ ((A ∩ B) ∩ (C ∩ C)) = ((A ∩ C) ∩ (B ∩ C))
4	2, 3	eqtr3 1121	1 ⊢ ((A ∩ B) ∩ C) = ((A ∩ C) ∩ (B ∩ C))

Syntax hints:   = wceq 1091   ∩ cin 1486

This theorem is referenced by:  difindir 1684

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-in 1491

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