Proof: Distribute Subtract2
Let's prove the following theorem:
(c ⋅ a) - (c ⋅ b) = c ⋅ (a - b)
Proof:
| # | Claim | Reason |
|---|---|---|
| 1 | (a - b) ⋅ c = (a ⋅ c) - (b ⋅ c) | (a - b) ⋅ c = (a ⋅ c) - (b ⋅ c) |
| 2 | c ⋅ a = a ⋅ c | c ⋅ a = a ⋅ c |
| 3 | (a - b) ⋅ c = (c ⋅ a) - (b ⋅ c) | if c ⋅ a = a ⋅ c and (a - b) ⋅ c = (a ⋅ c) - (b ⋅ c), then (a - b) ⋅ c = (c ⋅ a) - (b ⋅ c) |
| 4 | b ⋅ c = c ⋅ b | b ⋅ c = c ⋅ b |
| 5 | (a - b) ⋅ c = (c ⋅ a) - (c ⋅ b) | if b ⋅ c = c ⋅ b and (a - b) ⋅ c = (c ⋅ a) - (b ⋅ c), then (a - b) ⋅ c = (c ⋅ a) - (c ⋅ b) |
| 6 | (a - b) ⋅ c = c ⋅ (a - b) | (a - b) ⋅ c = c ⋅ (a - b) |
| 7 | c ⋅ (a - b) = (c ⋅ a) - (c ⋅ b) | if (a - b) ⋅ c = c ⋅ (a - b) and (a - b) ⋅ c = (c ⋅ a) - (c ⋅ b), then c ⋅ (a - b) = (c ⋅ a) - (c ⋅ b) |
| 8 | (c ⋅ a) - (c ⋅ b) = c ⋅ (a - b) | if c ⋅ (a - b) = (c ⋅ a) - (c ⋅ b), then (c ⋅ a) - (c ⋅ b) = c ⋅ (a - b) |
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