Theorem sumeq1d

Description: Equality deduction for sum. (Contributed by NM, 1-Nov-2005)

	Ref	Expression
Hypothesis	sumeq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	sumeq1d	⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 𝐶 = Σ 𝑘 ∈ 𝐵 𝐶 )

Proof

Step	Hyp	Ref	Expression
1		sumeq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		sumeq1	⊢ ( 𝐴 = 𝐵 → Σ 𝑘 ∈ 𝐴 𝐶 = Σ 𝑘 ∈ 𝐵 𝐶 )
3	1 2	syl	⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 𝐶 = Σ 𝑘 ∈ 𝐵 𝐶 )