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 + b) + (c + d) = (a + (c + d)) + b and (a + (c + d)) + b = ((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 (b + c) + a = 0 + a and 0 + a = 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 ((a + c) + d) + b = ((b + c) + a) + d and ((b + c) + a) + d = a + d, then ((a + c) + d) + b = a + d |
11 | (a + b) + (c + d) = a + d | if (a + b) + (c + d) = ((a + c) + d) + b and ((a + c) + d) + b = a + d, then (a + b) + (c + d) = a + d |
Comments
Please log in to add comments