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Metamath Proof Explorer
Description: Equality deduction for sum. (Contributed by NM, 3-Jan-2006) (Revised by Mario Carneiro, 31-Jan-2014)
|
|
Ref |
Expression |
|
Hypothesis |
sumeq2dv.1 |
⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐵 = 𝐶 ) |
|
Assertion |
sumeq2dv |
⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 𝐵 = Σ 𝑘 ∈ 𝐴 𝐶 ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
sumeq2dv.1 |
⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐵 = 𝐶 ) |
| 2 |
1
|
ralrimiva |
⊢ ( 𝜑 → ∀ 𝑘 ∈ 𝐴 𝐵 = 𝐶 ) |
| 3 |
2
|
sumeq2d |
⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 𝐵 = Σ 𝑘 ∈ 𝐴 𝐶 ) |