Metamath Proof Explorer


Theorem prodeq2d

Description: Equality deduction for product. Note that unlike prodeq2dv , k may occur in ph . (Contributed by Scott Fenton, 4-Dec-2017)

Ref Expression
Hypothesis prodeq2d.1 ⊢ ( 𝜑 → ∀ 𝑘 ∈ 𝐴 𝐵 = 𝐶 )
Assertion prodeq2d ( 𝜑 → ∏ 𝑘 ∈ 𝐴 𝐵 = ∏ 𝑘 ∈ 𝐴 𝐶 )

Proof

Step Hyp Ref Expression
1 prodeq2d.1 ⊢ ( 𝜑 → ∀ 𝑘 ∈ 𝐴 𝐵 = 𝐶 )
2 prodeq2 ⊢ ( ∀ 𝑘 ∈ 𝐴 𝐵 = 𝐶 → ∏ 𝑘 ∈ 𝐴 𝐵 = ∏ 𝑘 ∈ 𝐴 𝐶 )
3 1 2 syl ⊢ ( 𝜑 → ∏ 𝑘 ∈ 𝐴 𝐵 = ∏ 𝑘 ∈ 𝐴 𝐶 )