Metamath Proof Explorer


Theorem ssdifsn

Description: Subset of a set with an element removed. (Contributed by Emmett Weisz, 7-Jul-2021) (Proof shortened by JJ, 31-May-2022)

Ref Expression
Assertion ssdifsn A B C A B ¬ C A

Proof

Step Hyp Ref Expression
1 difss2 A B C A B
2 reldisj A B A C = A B C
3 2 bicomd A B A B C A C =
4 1 3 biadanii A B C A B A C =
5 disjsn A C = ¬ C A
6 5 anbi2i A B A C = A B ¬ C A
7 4 6 bitri A B C A B ¬ C A