Theorem 2cprodeq2dv

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

	Ref	Expression
Hypothesis	2cprodeq2dv.1	$$⊢\left(\phi \wedge j\in A\wedge k\in B\right)\to C=D$$
Assertion	2cprodeq2dv	$$⊢\phi \to \prod _{j\in A}\prod _{k\in B}C=\prod _{j\in A}\prod _{k\in B}D$$

Proof

Step	Hyp	Ref	Expression
1		2cprodeq2dv.1	$$⊢\left(\phi \wedge j\in A\wedge k\in B\right)\to C=D$$
2	1	3expa	$$⊢\left(\left(\phi \wedge j\in A\right)\wedge k\in B\right)\to C=D$$
3	2	prodeq2dv	$$⊢\left(\phi \wedge j\in A\right)\to \prod _{k\in B}C=\prod _{k\in B}D$$
4	3	prodeq2dv	$$⊢\phi \to \prod _{j\in A}\prod _{k\in B}C=\prod _{j\in A}\prod _{k\in B}D$$