Proof: Equation
Let's prove the following theorem:
if b + c = 0, then (a + b) + (c + d) = a + d
Proof:
Given
| 1 | b + c = 0 |
|---|
| # | Claim | Reason |
|---|---|---|
| 1 | (a + b) + (c + d) = (a + (c + d)) + b | (a + b) + (c + d) = (a + (c + d)) + b |
| 2 | a + (c + d) = (a + c) + d | a + (c + d) = (a + c) + d |
| 3 | (a + (c + d)) + b = ((a + c) + d) + b | if a + (c + d) = (a + c) + d, then (a + (c + d)) + b = ((a + c) + d) + b |
| 4 | (a + b) + (c + d) = ((a + c) + d) + b | if (a + (c + d)) + b = ((a + c) + d) + b and (a + b) + (c + d) = (a + (c + d)) + b, then (a + b) + (c + d) = ((a + c) + d) + b |
| 5 | ((a + c) + d) + b = ((b + c) + a) + d | ((a + c) + d) + b = ((b + c) + a) + d |
| 6 | (b + c) + a = 0 + a | if b + c = 0, then (b + c) + a = 0 + a |
| 7 | 0 + a = a | 0 + a = a |
| 8 | (b + c) + a = a | if 0 + a = a and (b + c) + a = 0 + a, then (b + c) + a = a |
| 9 | ((b + c) + a) + d = a + d | if (b + c) + a = a, then ((b + c) + a) + d = a + d |
| 10 | ((a + c) + d) + b = a + d | if ((b + c) + a) + d = a + d and ((a + c) + d) + b = ((b + c) + a) + d, then ((a + c) + d) + b = a + d |
| 11 | (a + b) + (c + d) = a + d | if ((a + c) + d) + b = a + d and (a + b) + (c + d) = ((a + c) + d) + b, then (a + b) + (c + d) = a + d |
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