Metamath Proof Explorer


Theorem inssdif0

Description: Intersection, subclass, and difference relationship. (Contributed by NM, 27-Oct-1996) (Proof shortened by Andrew Salmon, 26-Jun-2011) (Proof shortened by Wolf Lammen, 30-Sep-2014) (Proof shortened by BJ, 18-Jul-2026)

Ref Expression
Assertion inssdif0 A B C A B C =

Proof

Step Hyp Ref Expression
1 ssdif0 A B C A B C =
2 indif2 A B C = A B C
3 2 eqeq1i A B C = A B C =
4 1 3 bitr4i A B C A B C =