Metamath Proof Explorer


Theorem in12

Description: A rearrangement of intersection. (Contributed by NM, 21-Apr-2001)

Ref Expression
Assertion in12 ABC=BAC

Proof

Step Hyp Ref Expression
1 incom AB=BA
2 1 ineq1i ABC=BAC
3 inass ABC=ABC
4 inass BAC=BAC
5 2 3 4 3eqtr3i ABC=BAC