Theorem supeq1d

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

	Ref	Expression
Hypothesis	supeq1d.1	⊢ ( 𝜑 → 𝐵 = 𝐶 )
Assertion	supeq1d	⊢ ( 𝜑 → sup ( 𝐵 , 𝐴 , 𝑅 ) = sup ( 𝐶 , 𝐴 , 𝑅 ) )

Proof

Step	Hyp	Ref	Expression
1		supeq1d.1	⊢ ( 𝜑 → 𝐵 = 𝐶 )
2		supeq1	⊢ ( 𝐵 = 𝐶 → sup ( 𝐵 , 𝐴 , 𝑅 ) = sup ( 𝐶 , 𝐴 , 𝑅 ) )
3	1 2	syl	⊢ ( 𝜑 → sup ( 𝐵 , 𝐴 , 𝑅 ) = sup ( 𝐶 , 𝐴 , 𝑅 ) )