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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