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 neneq
 |-  ( A =/= C -> -. A = C )
2 1 adantl
 |-  ( ( A e. { B , C } /\ A =/= C ) -> -. A = C )
3 elpri
 |-  ( A e. { B , C } -> ( A = B \/ A = C ) )
4 3 adantr
 |-  ( ( A e. { B , C } /\ A =/= C ) -> ( A = B \/ A = C ) )
5 orcom
 |-  ( ( A = B \/ A = C ) <-> ( A = C \/ A = B ) )
6 df-or
 |-  ( ( A = C \/ A = B ) <-> ( -. A = C -> A = B ) )
7 5 6 bitri
 |-  ( ( A = B \/ A = C ) <-> ( -. A = C -> A = B ) )
8 4 7 sylib
 |-  ( ( A e. { B , C } /\ A =/= C ) -> ( -. A = C -> A = B ) )
9 2 8 mpd
 |-  ( ( A e. { B , C } /\ A =/= C ) -> A = B )