Metamath Proof Explorer


Theorem supeq1i

Description: Equality inference for supremum. (Contributed by Paul Chapman, 22-Jun-2011)

Ref Expression
Hypothesis supeq1i.1
|- B = C
Assertion supeq1i
|- sup ( B , A , R ) = sup ( C , A , R )

Proof

Step Hyp Ref Expression
1 supeq1i.1
 |-  B = C
2 supeq1
 |-  ( B = C -> sup ( B , A , R ) = sup ( C , A , R ) )
3 1 2 ax-mp
 |-  sup ( B , A , R ) = sup ( C , A , R )