Proof: Subtract Both Sides 2
Let's prove the following theorem:
if a = b + c, then a + (b ⋅ (-1)) = c
Proof:
Given
| 1 | a = b + c |
|---|
| # | Claim | Reason |
|---|---|---|
| 1 | b + c = c + b | b + c = c + b |
| 2 | a = c + b | if a = b + c and b + c = c + b, then a = c + b |
| 3 | a + (b ⋅ (-1)) = (c + b) + (b ⋅ (-1)) | if a = c + b, then a + (b ⋅ (-1)) = (c + b) + (b ⋅ (-1)) |
| 4 | (c + b) + (b ⋅ (-1)) = c + (b + (b ⋅ (-1))) | (c + b) + (b ⋅ (-1)) = c + (b + (b ⋅ (-1))) |
| 5 | a + (b ⋅ (-1)) = c + (b + (b ⋅ (-1))) | if a + (b ⋅ (-1)) = (c + b) + (b ⋅ (-1)) and (c + b) + (b ⋅ (-1)) = c + (b + (b ⋅ (-1))), then a + (b ⋅ (-1)) = c + (b + (b ⋅ (-1))) |
| 6 | b + (b ⋅ (-1)) = 0 | b + (b ⋅ (-1)) = 0 |
| 7 | c + (b + (b ⋅ (-1))) = c + 0 | if b + (b ⋅ (-1)) = 0, then c + (b + (b ⋅ (-1))) = c + 0 |
| 8 | a + (b ⋅ (-1)) = c + 0 | if a + (b ⋅ (-1)) = c + (b + (b ⋅ (-1))) and c + (b + (b ⋅ (-1))) = c + 0, then a + (b ⋅ (-1)) = c + 0 |
| 9 | c + 0 = c | c + 0 = c |
| 10 | a + (b ⋅ (-1)) = c | if a + (b ⋅ (-1)) = c + 0 and c + 0 = c, then a + (b ⋅ (-1)) = c |
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