Metamath Proof Explorer


Theorem infeq123d

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

Ref Expression
Hypotheses infeq123d.a ( 𝜑𝐴 = 𝐷 )
infeq123d.b ( 𝜑𝐵 = 𝐸 )
infeq123d.c ( 𝜑𝐶 = 𝐹 )
Assertion infeq123d ( 𝜑 → inf ( 𝐴 , 𝐵 , 𝐶 ) = inf ( 𝐷 , 𝐸 , 𝐹 ) )

Proof

Step Hyp Ref Expression
1 infeq123d.a ( 𝜑𝐴 = 𝐷 )
2 infeq123d.b ( 𝜑𝐵 = 𝐸 )
3 infeq123d.c ( 𝜑𝐶 = 𝐹 )
4 3 cnveqd ( 𝜑 𝐶 = 𝐹 )
5 1 2 4 supeq123d ( 𝜑 → sup ( 𝐴 , 𝐵 , 𝐶 ) = sup ( 𝐷 , 𝐸 , 𝐹 ) )
6 df-inf inf ( 𝐴 , 𝐵 , 𝐶 ) = sup ( 𝐴 , 𝐵 , 𝐶 )
7 df-inf inf ( 𝐷 , 𝐸 , 𝐹 ) = sup ( 𝐷 , 𝐸 , 𝐹 )
8 5 6 7 3eqtr4g ( 𝜑 → inf ( 𝐴 , 𝐵 , 𝐶 ) = inf ( 𝐷 , 𝐸 , 𝐹 ) )