Proof: Subtract Both Sides
Let's prove the following theorem:
if a = b + c, then a + (c ⋅ (-1)) = b
Proof:
Given
| 1 | a = b + c |
|---|
| # | Claim | Reason |
|---|---|---|
| 1 | a + (c ⋅ (-1)) = (b + c) + (c ⋅ (-1)) | if a = b + c, then a + (c ⋅ (-1)) = (b + c) + (c ⋅ (-1)) |
| 2 | (b + c) + (c ⋅ (-1)) = b + (c + (c ⋅ (-1))) | (b + c) + (c ⋅ (-1)) = b + (c + (c ⋅ (-1))) |
| 3 | a + (c ⋅ (-1)) = b + (c + (c ⋅ (-1))) | if a + (c ⋅ (-1)) = (b + c) + (c ⋅ (-1)) and (b + c) + (c ⋅ (-1)) = b + (c + (c ⋅ (-1))), then a + (c ⋅ (-1)) = b + (c + (c ⋅ (-1))) |
| 4 | c + (c ⋅ (-1)) = 0 | c + (c ⋅ (-1)) = 0 |
| 5 | b + (c + (c ⋅ (-1))) = b + 0 | if c + (c ⋅ (-1)) = 0, then b + (c + (c ⋅ (-1))) = b + 0 |
| 6 | a + (c ⋅ (-1)) = b + 0 | if a + (c ⋅ (-1)) = b + (c + (c ⋅ (-1))) and b + (c + (c ⋅ (-1))) = b + 0, then a + (c ⋅ (-1)) = b + 0 |
| 7 | b + 0 = b | b + 0 = b |
| 8 | a + (c ⋅ (-1)) = b | if a + (c ⋅ (-1)) = b + 0 and b + 0 = b, then a + (c ⋅ (-1)) = b |
Comments
Please log in to add comments