Metamath Proof Explorer


Theorem eqtrb

Description: A transposition of equality. (Contributed by Thierry Arnoux, 20-Aug-2023)

Ref Expression
Assertion eqtrb ( ( 𝐴 = 𝐵𝐴 = 𝐶 ) ↔ ( 𝐴 = 𝐵𝐵 = 𝐶 ) )

Proof

Step Hyp Ref Expression
1 simpl ( ( 𝐴 = 𝐵𝐴 = 𝐶 ) → 𝐴 = 𝐵 )
2 eqtr2 ( ( 𝐴 = 𝐵𝐴 = 𝐶 ) → 𝐵 = 𝐶 )
3 1 2 jca ( ( 𝐴 = 𝐵𝐴 = 𝐶 ) → ( 𝐴 = 𝐵𝐵 = 𝐶 ) )
4 simpl ( ( 𝐴 = 𝐵𝐵 = 𝐶 ) → 𝐴 = 𝐵 )
5 eqtr ( ( 𝐴 = 𝐵𝐵 = 𝐶 ) → 𝐴 = 𝐶 )
6 4 5 jca ( ( 𝐴 = 𝐵𝐵 = 𝐶 ) → ( 𝐴 = 𝐵𝐴 = 𝐶 ) )
7 3 6 impbii ( ( 𝐴 = 𝐵𝐴 = 𝐶 ) ↔ ( 𝐴 = 𝐵𝐵 = 𝐶 ) )