Proof: Equation

Let's prove the following theorem:

if b + c = 0, then (a + b) + (c + d) = a + d

Proof:

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Given
1 b + c = 0
Proof Table
# 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
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