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