Description: Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017)
Ref | Expression | ||
---|---|---|---|
Hypotheses | prodeq12dv.1 | ⊢ ( 𝜑 → 𝐴 = 𝐵 ) | |
prodeq12dv.2 | ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐶 = 𝐷 ) | ||
Assertion | prodeq12dv | ⊢ ( 𝜑 → ∏ 𝑘 ∈ 𝐴 𝐶 = ∏ 𝑘 ∈ 𝐵 𝐷 ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | prodeq12dv.1 | ⊢ ( 𝜑 → 𝐴 = 𝐵 ) | |
2 | prodeq12dv.2 | ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐶 = 𝐷 ) | |
3 | 2 | prodeq2dv | ⊢ ( 𝜑 → ∏ 𝑘 ∈ 𝐴 𝐶 = ∏ 𝑘 ∈ 𝐴 𝐷 ) |
4 | 1 | prodeq1d | ⊢ ( 𝜑 → ∏ 𝑘 ∈ 𝐴 𝐷 = ∏ 𝑘 ∈ 𝐵 𝐷 ) |
5 | 3 4 | eqtrd | ⊢ ( 𝜑 → ∏ 𝑘 ∈ 𝐴 𝐶 = ∏ 𝑘 ∈ 𝐵 𝐷 ) |