Metamath Proof Explorer


Theorem infeq2

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

Ref Expression
Assertion infeq2
|- ( B = C -> inf ( A , B , R ) = inf ( A , C , R ) )

Proof

Step Hyp Ref Expression
1 supeq2
 |-  ( B = C -> sup ( A , B , `' R ) = sup ( A , C , `' R ) )
2 df-inf
 |-  inf ( A , B , R ) = sup ( A , B , `' R )
3 df-inf
 |-  inf ( A , C , R ) = sup ( A , C , `' R )
4 1 2 3 3eqtr4g
 |-  ( B = C -> inf ( A , B , R ) = inf ( A , C , R ) )