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 ) )