Metamath Proof Explorer


Theorem infeq1d

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

Ref Expression
Hypothesis infeq1d.1 φ B = C
Assertion infeq1d φ sup B A R = sup C A R

Proof

Step Hyp Ref Expression
1 infeq1d.1 φ B = C
2 infeq1 B = C sup B A R = sup C A R
3 1 2 syl φ sup B A R = sup C A R