Proof: Divide Simplify 2
Let's prove the following theorem:
if not (d = 0), then (b ⋅ d) ⋅ (c / d) = b ⋅ c
Proof:
Given
| 1 | not (d = 0) |
|---|
| # | Claim | Reason |
|---|---|---|
| 1 | (b ⋅ d) ⋅ c = (b ⋅ c) ⋅ d | (b ⋅ d) ⋅ c = (b ⋅ c) ⋅ d |
| 2 | ((b ⋅ d) ⋅ c) / d = ((b ⋅ c) ⋅ d) / d | if (b ⋅ d) ⋅ c = (b ⋅ c) ⋅ d, then ((b ⋅ d) ⋅ c) / d = ((b ⋅ c) ⋅ d) / d |
| 3 | d / d = 1 | if not (d = 0), then d / d = 1 |
| 4 | ((b ⋅ c) ⋅ d) / d = (b ⋅ c) ⋅ (d / d) | ((b ⋅ c) ⋅ d) / d = (b ⋅ c) ⋅ (d / d) |
| 5 | ((b ⋅ c) ⋅ d) / d = (b ⋅ c) ⋅ 1 | if d / d = 1 and ((b ⋅ c) ⋅ d) / d = (b ⋅ c) ⋅ (d / d), then ((b ⋅ c) ⋅ d) / d = (b ⋅ c) ⋅ 1 |
| 6 | (b ⋅ c) ⋅ 1 = b ⋅ c | (b ⋅ c) ⋅ 1 = b ⋅ c |
| 7 | ((b ⋅ c) ⋅ d) / d = b ⋅ c | if (b ⋅ c) ⋅ 1 = b ⋅ c and ((b ⋅ c) ⋅ d) / d = (b ⋅ c) ⋅ 1, then ((b ⋅ c) ⋅ d) / d = b ⋅ c |
| 8 | ((b ⋅ d) ⋅ c) / d = b ⋅ c | if ((b ⋅ c) ⋅ d) / d = b ⋅ c and ((b ⋅ d) ⋅ c) / d = ((b ⋅ c) ⋅ d) / d, then ((b ⋅ d) ⋅ c) / d = b ⋅ c |
| 9 | ((b ⋅ d) ⋅ c) / d = (b ⋅ d) ⋅ (c / d) | ((b ⋅ d) ⋅ c) / d = (b ⋅ d) ⋅ (c / d) |
| 10 | (b ⋅ d) ⋅ (c / d) = b ⋅ c | if ((b ⋅ d) ⋅ c) / d = b ⋅ c and ((b ⋅ d) ⋅ c) / d = (b ⋅ d) ⋅ (c / d), then (b ⋅ d) ⋅ (c / d) = b ⋅ c |
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