Metamath Proof Explorer


Theorem prodeq12rdv

Description: Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017)

Ref Expression
Hypotheses prodeq12rdv.1 ( 𝜑𝐴 = 𝐵 )
prodeq12rdv.2 ( ( 𝜑𝑘𝐵 ) → 𝐶 = 𝐷 )
Assertion prodeq12rdv ( 𝜑 → ∏ 𝑘𝐴 𝐶 = ∏ 𝑘𝐵 𝐷 )

Proof

Step Hyp Ref Expression
1 prodeq12rdv.1 ( 𝜑𝐴 = 𝐵 )
2 prodeq12rdv.2 ( ( 𝜑𝑘𝐵 ) → 𝐶 = 𝐷 )
3 1 prodeq1d ( 𝜑 → ∏ 𝑘𝐴 𝐶 = ∏ 𝑘𝐵 𝐶 )
4 2 prodeq2dv ( 𝜑 → ∏ 𝑘𝐵 𝐶 = ∏ 𝑘𝐵 𝐷 )
5 3 4 eqtrd ( 𝜑 → ∏ 𝑘𝐴 𝐶 = ∏ 𝑘𝐵 𝐷 )