Metamath Proof Explorer


Theorem elprn2

Description: A member of an unordered pair that is not the "second", must be the "first". (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Assertion elprn2
|- ( ( A e. { B , C } /\ A =/= C ) -> A = B )

Proof

Step Hyp Ref Expression
1 elpri
 |-  ( A e. { B , C } -> ( A = B \/ A = C ) )
2 1 adantr
 |-  ( ( A e. { B , C } /\ A =/= C ) -> ( A = B \/ A = C ) )
3 neneq
 |-  ( A =/= C -> -. A = C )
4 3 adantl
 |-  ( ( A e. { B , C } /\ A =/= C ) -> -. A = C )
5 2 4 olcnd
 |-  ( ( A e. { B , C } /\ A =/= C ) -> A = B )