Proof: Reorder Terms
Let's prove the following theorem:
((a ⋅ b) ⋅ c) ⋅ d = ((b ⋅ d) ⋅ a) ⋅ c
Proof:
| # | Claim | Reason |
|---|---|---|
| 1 | ((a ⋅ b) ⋅ c) ⋅ d = (a ⋅ b) ⋅ (c ⋅ d) | ((a ⋅ b) ⋅ c) ⋅ d = (a ⋅ b) ⋅ (c ⋅ d) |
| 2 | c ⋅ d = d ⋅ c | c ⋅ d = d ⋅ c |
| 3 | (a ⋅ b) ⋅ (c ⋅ d) = (a ⋅ b) ⋅ (d ⋅ c) | if c ⋅ d = d ⋅ c, then (a ⋅ b) ⋅ (c ⋅ d) = (a ⋅ b) ⋅ (d ⋅ c) |
| 4 | a ⋅ b = b ⋅ a | a ⋅ b = b ⋅ a |
| 5 | (a ⋅ b) ⋅ (c ⋅ d) = (b ⋅ a) ⋅ (d ⋅ c) | if (a ⋅ b) ⋅ (c ⋅ d) = (a ⋅ b) ⋅ (d ⋅ c) and a ⋅ b = b ⋅ a, then (a ⋅ b) ⋅ (c ⋅ d) = (b ⋅ a) ⋅ (d ⋅ c) |
| 6 | (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ a) ⋅ d) ⋅ c | (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ a) ⋅ d) ⋅ c |
| 7 | (b ⋅ a) ⋅ d = (b ⋅ d) ⋅ a | (b ⋅ a) ⋅ d = (b ⋅ d) ⋅ a |
| 8 | ((b ⋅ a) ⋅ d) ⋅ c = ((b ⋅ d) ⋅ a) ⋅ c | if (b ⋅ a) ⋅ d = (b ⋅ d) ⋅ a, then ((b ⋅ a) ⋅ d) ⋅ c = ((b ⋅ d) ⋅ a) ⋅ c |
| 9 | (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ d) ⋅ a) ⋅ c | if (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ a) ⋅ d) ⋅ c and ((b ⋅ a) ⋅ d) ⋅ c = ((b ⋅ d) ⋅ a) ⋅ c, then (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ d) ⋅ a) ⋅ c |
| 10 | (a ⋅ b) ⋅ (c ⋅ d) = ((b ⋅ d) ⋅ a) ⋅ c | if (a ⋅ b) ⋅ (c ⋅ d) = (b ⋅ a) ⋅ (d ⋅ c) and (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ d) ⋅ a) ⋅ c, then (a ⋅ b) ⋅ (c ⋅ d) = ((b ⋅ d) ⋅ a) ⋅ c |
| 11 | ((a ⋅ b) ⋅ c) ⋅ d = ((b ⋅ d) ⋅ a) ⋅ c | if ((a ⋅ b) ⋅ c) ⋅ d = (a ⋅ b) ⋅ (c ⋅ d) and (a ⋅ b) ⋅ (c ⋅ d) = ((b ⋅ d) ⋅ a) ⋅ c, then ((a ⋅ b) ⋅ c) ⋅ d = ((b ⋅ d) ⋅ a) ⋅ c |
Comments
Please log in to add comments