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Metamath Proof Explorer

Theorem 2cprodeq2dv

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

		Ref	Expression
	Hypothesis	2cprodeq2dv.1	$⊢ (φ \land j \in A \land k \in B) \to C = D$
	Assertion	2cprodeq2dv	$⊢ φ \to \prod_{j \in A} \prod_{k \in B} C = \prod_{j \in A} \prod_{k \in B} D$

Step	Hyp	Ref	Expression
1		2cprodeq2dv.1	$⊢ (φ \land j \in A \land k \in B) \to C = D$
2	1	3expa	$⊢ ((φ \land j \in A) \land k \in B) \to C = D$
3	2	prodeq2dv	$⊢ (φ \land j \in A) \to \prod_{k \in B} C = \prod_{k \in B} D$
4	3	prodeq2dv	$⊢ φ \to \prod_{j \in A} \prod_{k \in B} C = \prod_{j \in A} \prod_{k \in B} D$