Metamath Proof Explorer


Theorem infeq3

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

Ref Expression
Assertion infeq3 ( 𝑅 = 𝑆 → inf ( 𝐴 , 𝐵 , 𝑅 ) = inf ( 𝐴 , 𝐵 , 𝑆 ) )

Proof

Step Hyp Ref Expression
1 cnveq ( 𝑅 = 𝑆 𝑅 = 𝑆 )
2 supeq3 ( 𝑅 = 𝑆 → sup ( 𝐴 , 𝐵 , 𝑅 ) = sup ( 𝐴 , 𝐵 , 𝑆 ) )
3 1 2 syl ( 𝑅 = 𝑆 → sup ( 𝐴 , 𝐵 , 𝑅 ) = sup ( 𝐴 , 𝐵 , 𝑆 ) )
4 df-inf inf ( 𝐴 , 𝐵 , 𝑅 ) = sup ( 𝐴 , 𝐵 , 𝑅 )
5 df-inf inf ( 𝐴 , 𝐵 , 𝑆 ) = sup ( 𝐴 , 𝐵 , 𝑆 )
6 3 4 5 3eqtr4g ( 𝑅 = 𝑆 → inf ( 𝐴 , 𝐵 , 𝑅 ) = inf ( 𝐴 , 𝐵 , 𝑆 ) )