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Metamath Proof Explorer
Description: Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017)
|
|
Ref |
Expression |
|
Hypothesis |
prodeq2dv.1 |
⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐵 = 𝐶 ) |
|
Assertion |
prodeq2dv |
⊢ ( 𝜑 → ∏ 𝑘 ∈ 𝐴 𝐵 = ∏ 𝑘 ∈ 𝐴 𝐶 ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
prodeq2dv.1 |
⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐵 = 𝐶 ) |
| 2 |
1
|
ralrimiva |
⊢ ( 𝜑 → ∀ 𝑘 ∈ 𝐴 𝐵 = 𝐶 ) |
| 3 |
2
|
prodeq2d |
⊢ ( 𝜑 → ∏ 𝑘 ∈ 𝐴 𝐵 = ∏ 𝑘 ∈ 𝐴 𝐶 ) |