Metamath Proof Explorer


Theorem inindi

Description: Intersection distributes over itself. (Contributed by NM, 6-May-1994)

Ref Expression
Assertion inindi A B C = A B A C

Proof

Step Hyp Ref Expression
1 inidm A A = A
2 1 ineq1i A A B C = A B C
3 in4 A A B C = A B A C
4 2 3 eqtr3i A B C = A B A C