Proof: Algebra 6
Let's prove the following theorem:
(a / b) ⋅ d = (d / b) ⋅ a
    
    
    
    Proof:
| # | 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 ⋅ a) ⋅ (1 / b) = (d ⋅ (1 / b)) ⋅ a and (d ⋅ (1 / b)) ⋅ a = (d / b) ⋅ a, then (d ⋅ a) ⋅ (1 / b) = (d / b) ⋅ a | 
| 8 | (d ⋅ a) / b = (d / b) ⋅ a | if (d ⋅ a) / b = (d ⋅ a) ⋅ (1 / b) and (d ⋅ a) ⋅ (1 / b) = (d / b) ⋅ a, then (d ⋅ a) / b = (d / b) ⋅ a | 
| 9 | d ⋅ (a / b) = (d / b) ⋅ a | if d ⋅ (a / b) = (d ⋅ a) / b and (d ⋅ a) / b = (d / b) ⋅ a, then d ⋅ (a / b) = (d / b) ⋅ a | 
| 10 | (a / b) ⋅ d = (d / b) ⋅ a | if (a / b) ⋅ d = d ⋅ (a / b) and d ⋅ (a / b) = (d / b) ⋅ a, then (a / b) ⋅ d = (d / b) ⋅ a | 
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