Theorem 2cprodeq2dv

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

	Ref	Expression
Hypothesis	2cprodeq2dv.1	⊢ ( ( 𝜑 ∧ 𝑗 ∈ 𝐴 ∧ 𝑘 ∈ 𝐵 ) → 𝐶 = 𝐷 )
Assertion	2cprodeq2dv	⊢ ( 𝜑 → ∏ 𝑗 ∈ 𝐴 ∏ 𝑘 ∈ 𝐵 𝐶 = ∏ 𝑗 ∈ 𝐴 ∏ 𝑘 ∈ 𝐵 𝐷 )

Proof

Step	Hyp	Ref	Expression
1		2cprodeq2dv.1	⊢ ( ( 𝜑 ∧ 𝑗 ∈ 𝐴 ∧ 𝑘 ∈ 𝐵 ) → 𝐶 = 𝐷 )
2	1	3expa	⊢ ( ( ( 𝜑 ∧ 𝑗 ∈ 𝐴 ) ∧ 𝑘 ∈ 𝐵 ) → 𝐶 = 𝐷 )
3	2	prodeq2dv	⊢ ( ( 𝜑 ∧ 𝑗 ∈ 𝐴 ) → ∏ 𝑘 ∈ 𝐵 𝐶 = ∏ 𝑘 ∈ 𝐵 𝐷 )
4	3	prodeq2dv	⊢ ( 𝜑 → ∏ 𝑗 ∈ 𝐴 ∏ 𝑘 ∈ 𝐵 𝐶 = ∏ 𝑗 ∈ 𝐴 ∏ 𝑘 ∈ 𝐵 𝐷 )