Metamath Proof Explorer


Theorem neldifpr1

Description: The first element of a pair is not an element of a difference with this pair. (Contributed by Thierry Arnoux, 20-Nov-2023)

Ref Expression
Assertion neldifpr1
|- -. A e. ( C \ { A , B } )

Proof

Step Hyp Ref Expression
1 neirr
 |-  -. A =/= A
2 eldifpr
 |-  ( A e. ( C \ { A , B } ) <-> ( A e. C /\ A =/= A /\ A =/= B ) )
3 2 simp2bi
 |-  ( A e. ( C \ { A , B } ) -> A =/= A )
4 1 3 mto
 |-  -. A e. ( C \ { A , B } )