Metamath Proof Explorer


Theorem infeq1d

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

Ref Expression
Hypothesis infeq1d.1 ( 𝜑𝐵 = 𝐶 )
Assertion infeq1d ( 𝜑 → inf ( 𝐵 , 𝐴 , 𝑅 ) = inf ( 𝐶 , 𝐴 , 𝑅 ) )

Proof

Step Hyp Ref Expression
1 infeq1d.1 ( 𝜑𝐵 = 𝐶 )
2 infeq1 ( 𝐵 = 𝐶 → inf ( 𝐵 , 𝐴 , 𝑅 ) = inf ( 𝐶 , 𝐴 , 𝑅 ) )
3 1 2 syl ( 𝜑 → inf ( 𝐵 , 𝐴 , 𝑅 ) = inf ( 𝐶 , 𝐴 , 𝑅 ) )