Proof: Simplify Product 2

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	(a ⋅ c) ⋅ ((1 / b) ⋅ (1 / d)) = ((a ⋅ (1 / b)) ⋅ c) ⋅ (1 / d)	(a ⋅ c) ⋅ ((1 / b) ⋅ (1 / d)) = ((a ⋅ (1 / b)) ⋅ c) ⋅ (1 / d) (reorder_terms_8)
2	((a ⋅ (1 / b)) ⋅ c) ⋅ (1 / d) = (a ⋅ (1 / b)) ⋅ (c ⋅ (1 / d))	((a ⋅ (1 / b)) ⋅ c) ⋅ (1 / d) = (a ⋅ (1 / b)) ⋅ (c ⋅ (1 / d)) (Associative Property of Multiplication)
3	(a ⋅ (1 / b)) ⋅ (c ⋅ (1 / d)) = (a / b) ⋅ (c / d)	(a ⋅ (1 / b)) ⋅ (c ⋅ (1 / d)) = (a / b) ⋅ (c / d) (simplify_product)
4	(a ⋅ c) ⋅ ((1 / b) ⋅ (1 / d)) = (a / b) ⋅ (c / d)	if (a ⋅ (1 / b)) ⋅ (c ⋅ (1 / d)) = (a / b) ⋅ (c / d) and ((a ⋅ (1 / b)) ⋅ c) ⋅ (1 / d) = (a ⋅ (1 / b)) ⋅ (c ⋅ (1 / d)) and (a ⋅ c) ⋅ ((1 / b) ⋅ (1 / d)) = ((a ⋅ (1 / b)) ⋅ c) ⋅ (1 / d), then (a ⋅ c) ⋅ ((1 / b) ⋅ (1 / d)) = (a / b) ⋅ (c / d) (Transitive Property Application 1)

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