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 ∈ { 𝐵 , 𝐶 } ∧ 𝐴𝐶 ) → 𝐴 = 𝐵 )

Proof

Step Hyp Ref Expression
1 neneq ( 𝐴𝐶 → ¬ 𝐴 = 𝐶 )
2 1 adantl ( ( 𝐴 ∈ { 𝐵 , 𝐶 } ∧ 𝐴𝐶 ) → ¬ 𝐴 = 𝐶 )
3 elpri ( 𝐴 ∈ { 𝐵 , 𝐶 } → ( 𝐴 = 𝐵𝐴 = 𝐶 ) )
4 3 adantr ( ( 𝐴 ∈ { 𝐵 , 𝐶 } ∧ 𝐴𝐶 ) → ( 𝐴 = 𝐵𝐴 = 𝐶 ) )
5 orcom ( ( 𝐴 = 𝐵𝐴 = 𝐶 ) ↔ ( 𝐴 = 𝐶𝐴 = 𝐵 ) )
6 df-or ( ( 𝐴 = 𝐶𝐴 = 𝐵 ) ↔ ( ¬ 𝐴 = 𝐶𝐴 = 𝐵 ) )
7 5 6 bitri ( ( 𝐴 = 𝐵𝐴 = 𝐶 ) ↔ ( ¬ 𝐴 = 𝐶𝐴 = 𝐵 ) )
8 4 7 sylib ( ( 𝐴 ∈ { 𝐵 , 𝐶 } ∧ 𝐴𝐶 ) → ( ¬ 𝐴 = 𝐶𝐴 = 𝐵 ) )
9 2 8 mpd ( ( 𝐴 ∈ { 𝐵 , 𝐶 } ∧ 𝐴𝐶 ) → 𝐴 = 𝐵 )