Metamath Proof Explorer


Theorem elprn1

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

Ref Expression
Assertion elprn1
|- ( ( A e. { B , C } /\ A =/= B ) -> A = C )

Proof

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