Proof: Algebra 17b
Let's prove the following theorem:
(s / 2) / s = 1 / 2
    
    
    
    Proof:
| # | Claim | Reason | 
|---|---|---|
| 1 | s / 2 = s ⋅ (1 / 2) | s / 2 = s ⋅ (1 / 2) | 
| 2 | (s / 2) / s = (s ⋅ (1 / 2)) / s | if s / 2 = s ⋅ (1 / 2), then (s / 2) / s = (s ⋅ (1 / 2)) / s | 
| 3 | (s ⋅ (1 / 2)) / s = (s ⋅ (1 / 2)) ⋅ (1 / s) | (s ⋅ (1 / 2)) / s = (s ⋅ (1 / 2)) ⋅ (1 / s) | 
| 4 | (s ⋅ (1 / 2)) ⋅ (1 / s) = (s ⋅ (1 / s)) ⋅ (1 / 2) | (s ⋅ (1 / 2)) ⋅ (1 / s) = (s ⋅ (1 / s)) ⋅ (1 / 2) | 
| 5 | s ⋅ (1 / s) = 1 | s ⋅ (1 / s) = 1 | 
| 6 | (s ⋅ (1 / s)) ⋅ (1 / 2) = 1 ⋅ (1 / 2) | if s ⋅ (1 / s) = 1, then (s ⋅ (1 / s)) ⋅ (1 / 2) = 1 ⋅ (1 / 2) | 
| 7 | 1 ⋅ (1 / 2) = 1 / 2 | 1 ⋅ (1 / 2) = 1 / 2 | 
| 8 | (s ⋅ (1 / s)) ⋅ (1 / 2) = 1 / 2 | if (s ⋅ (1 / s)) ⋅ (1 / 2) = 1 ⋅ (1 / 2) and 1 ⋅ (1 / 2) = 1 / 2, then (s ⋅ (1 / s)) ⋅ (1 / 2) = 1 / 2 | 
| 9 | (s ⋅ (1 / 2)) ⋅ (1 / s) = 1 / 2 | if (s ⋅ (1 / 2)) ⋅ (1 / s) = (s ⋅ (1 / s)) ⋅ (1 / 2) and (s ⋅ (1 / s)) ⋅ (1 / 2) = 1 / 2, then (s ⋅ (1 / 2)) ⋅ (1 / s) = 1 / 2 | 
| 10 | (s ⋅ (1 / 2)) / s = 1 / 2 | if (s ⋅ (1 / 2)) / s = (s ⋅ (1 / 2)) ⋅ (1 / s) and (s ⋅ (1 / 2)) ⋅ (1 / s) = 1 / 2, then (s ⋅ (1 / 2)) / s = 1 / 2 | 
| 11 | (s / 2) / s = 1 / 2 | if (s / 2) / s = (s ⋅ (1 / 2)) / s and (s ⋅ (1 / 2)) / s = 1 / 2, then (s / 2) / s = 1 / 2 | 
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