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 ) )