Metamath Proof Explorer


Theorem inin

Description: Intersection with an intersection. (Contributed by Thierry Arnoux, 27-Dec-2016)

Ref Expression
Assertion inin AAB=AB

Proof

Step Hyp Ref Expression
1 in13 AAB=BAA
2 inidm AA=A
3 2 ineq2i BAA=BA
4 incom BA=AB
5 1 3 4 3eqtri AAB=AB