Proof: Divide Zero 2
Let's prove the following theorem:
if b > 0, then (c - c) / ((b + a) - a) = 0
Proof:
Given
| 1 | b > 0 |
|---|
| # | Claim | Reason |
|---|---|---|
| 1 | not (b = 0) | if b > 0, then not (b = 0) |
| 2 | c - c = 0 | c - c = 0 |
| 3 | (c - c) / b = 0 / b | if c - c = 0, then (c - c) / b = 0 / b |
| 4 | 0 / b = 0 | if not (b = 0), then 0 / b = 0 |
| 5 | (c - c) / b = 0 | if (c - c) / b = 0 / b and 0 / b = 0, then (c - c) / b = 0 |
| 6 | (b + a) - a = b | (b + a) - a = b |
| 7 | (c - c) / ((b + a) - a) = (c - c) / b | if (b + a) - a = b, then (c - c) / ((b + a) - a) = (c - c) / b |
| 8 | (c - c) / ((b + a) - a) = 0 | if (c - c) / ((b + a) - a) = (c - c) / b and (c - c) / b = 0, then (c - c) / ((b + a) - a) = 0 |
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