Proof: Reduction Example
Let's prove the following theorem:
if not (b = 0), then ((a / b) ⋅ c) ⋅ b = a ⋅ c
First we show that:
(a / b) ⋅ c ⋅ b = a ⋅ (1 / b) ⋅ c ⋅ b (step 3)
Then we use the Commutative and Associative properties to reorder some of the terms:
a ⋅ (1 / b)) ⋅ c ⋅ b = (1 / b) ⋅ b ⋅ a ⋅ c (step 4)
Then we show that:
(1 / b) ⋅ b ⋅ a ⋅ c = 1 ⋅ a ⋅ c (step 7)
Then we simplify the right side as follows:
1 ⋅ a ⋅ c = a ⋅ c (step 9)
Finally, we use the Transitive Property a few times to reach our conclusion.
Proof:
Given
| 1 | not (b = 0) |
|---|
| # | Claim | Reason |
|---|---|---|
| 1 | a / b = a ⋅ (1 / b) | a / b = a ⋅ (1 / b) |
| 2 | (a / b) ⋅ c = (a ⋅ (1 / b)) ⋅ c | if a / b = a ⋅ (1 / b), then (a / b) ⋅ c = (a ⋅ (1 / b)) ⋅ c |
| 3 | ((a / b) ⋅ c) ⋅ b = ((a ⋅ (1 / b)) ⋅ c) ⋅ b | if (a / b) ⋅ c = (a ⋅ (1 / b)) ⋅ c, then ((a / b) ⋅ c) ⋅ b = ((a ⋅ (1 / b)) ⋅ c) ⋅ b |
| 4 | ((a ⋅ (1 / b)) ⋅ c) ⋅ b = (((1 / b) ⋅ b) ⋅ a) ⋅ c | ((a ⋅ (1 / b)) ⋅ c) ⋅ b = (((1 / b) ⋅ b) ⋅ a) ⋅ c |
| 5 | (1 / b) ⋅ b = 1 | if not (b = 0), then (1 / b) ⋅ b = 1 |
| 6 | ((1 / b) ⋅ b) ⋅ a = 1 ⋅ a | if (1 / b) ⋅ b = 1, then ((1 / b) ⋅ b) ⋅ a = 1 ⋅ a |
| 7 | (((1 / b) ⋅ b) ⋅ a) ⋅ c = (1 ⋅ a) ⋅ c | if ((1 / b) ⋅ b) ⋅ a = 1 ⋅ a, then (((1 / b) ⋅ b) ⋅ a) ⋅ c = (1 ⋅ a) ⋅ c |
| 8 | 1 ⋅ a = a | 1 ⋅ a = a |
| 9 | (1 ⋅ a) ⋅ c = a ⋅ c | if 1 ⋅ a = a, then (1 ⋅ a) ⋅ c = a ⋅ c |
| 10 | (((1 / b) ⋅ b) ⋅ a) ⋅ c = a ⋅ c | if (((1 / b) ⋅ b) ⋅ a) ⋅ c = (1 ⋅ a) ⋅ c and (1 ⋅ a) ⋅ c = a ⋅ c, then (((1 / b) ⋅ b) ⋅ a) ⋅ c = a ⋅ c |
| 11 | ((a ⋅ (1 / b)) ⋅ c) ⋅ b = a ⋅ c | if ((a ⋅ (1 / b)) ⋅ c) ⋅ b = (((1 / b) ⋅ b) ⋅ a) ⋅ c and (((1 / b) ⋅ b) ⋅ a) ⋅ c = a ⋅ c, then ((a ⋅ (1 / b)) ⋅ c) ⋅ b = a ⋅ c |
| 12 | ((a / b) ⋅ c) ⋅ b = a ⋅ c | if ((a / b) ⋅ c) ⋅ b = ((a ⋅ (1 / b)) ⋅ c) ⋅ b and ((a ⋅ (1 / b)) ⋅ c) ⋅ b = a ⋅ c, then ((a / b) ⋅ c) ⋅ b = a ⋅ c |
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