Proof: Algebra 6

Let's prove the following theorem:

(a / b) ⋅ d = (d / b) ⋅ a

Proof:

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Proof Table
# Claim Reason
1 (a / b) ⋅ d = d ⋅ (a / b) (a / b) ⋅ d = d ⋅ (a / b)
2 d ⋅ (a / b) = (d ⋅ a) / b d ⋅ (a / b) = (d ⋅ a) / b
3 (d ⋅ a) / b = (d ⋅ a) ⋅ (1 / b) (d ⋅ a) / b = (d ⋅ a) ⋅ (1 / b)
4 (d ⋅ a) ⋅ (1 / b) = (d ⋅ (1 / b)) ⋅ a (d ⋅ a) ⋅ (1 / b) = (d ⋅ (1 / b)) ⋅ a
5 d ⋅ (1 / b) = d / b d ⋅ (1 / b) = d / b
6 (d ⋅ (1 / b)) ⋅ a = (d / b) ⋅ a if d ⋅ (1 / b) = d / b, then (d ⋅ (1 / b)) ⋅ a = (d / b) ⋅ a
7 (d ⋅ a) ⋅ (1 / b) = (d / b) ⋅ a if (d ⋅ (1 / b)) ⋅ a = (d / b) ⋅ a and (d ⋅ a) ⋅ (1 / b) = (d ⋅ (1 / b)) ⋅ a, then (d ⋅ a) ⋅ (1 / b) = (d / b) ⋅ a
8 (d ⋅ a) / b = (d / b) ⋅ a if (d ⋅ a) ⋅ (1 / b) = (d / b) ⋅ a and (d ⋅ a) / b = (d ⋅ a) ⋅ (1 / b), then (d ⋅ a) / b = (d / b) ⋅ a
9 d ⋅ (a / b) = (d / b) ⋅ a if (d ⋅ a) / b = (d / b) ⋅ a and d ⋅ (a / b) = (d ⋅ a) / b, then d ⋅ (a / b) = (d / b) ⋅ a
10 (a / b) ⋅ d = (d / b) ⋅ a if d ⋅ (a / b) = (d / b) ⋅ a and (a / b) ⋅ d = d ⋅ (a / b), then (a / b) ⋅ d = (d / b) ⋅ a

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