Metamath Proof Explorer


Theorem infeq1

Description: Equality theorem for infimum. (Contributed by AV, 2-Sep-2020)

Ref Expression
Assertion infeq1 ( 𝐵 = 𝐶 → inf ( 𝐵 , 𝐴 , 𝑅 ) = inf ( 𝐶 , 𝐴 , 𝑅 ) )

Proof

Step Hyp Ref Expression
1 supeq1 ( 𝐵 = 𝐶 → sup ( 𝐵 , 𝐴 , 𝑅 ) = sup ( 𝐶 , 𝐴 , 𝑅 ) )
2 df-inf inf ( 𝐵 , 𝐴 , 𝑅 ) = sup ( 𝐵 , 𝐴 , 𝑅 )
3 df-inf inf ( 𝐶 , 𝐴 , 𝑅 ) = sup ( 𝐶 , 𝐴 , 𝑅 )
4 1 2 3 3eqtr4g ( 𝐵 = 𝐶 → inf ( 𝐵 , 𝐴 , 𝑅 ) = inf ( 𝐶 , 𝐴 , 𝑅 ) )