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