Proof: Simplify 3

Let's prove the following theorem:

if not (2 = 0), then ((b2) + (a2)) / 2 = b + a

Proof:

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Given
1 not (2 = 0)
Proof Table
# Claim Reason
1 ((b2) + (a2)) / 2 = ((b2) + (a2)) ⋅ (1 / 2) ((b2) + (a2)) / 2 = ((b2) + (a2)) ⋅ (1 / 2)
2 ((b2) + (a2)) ⋅ (1 / 2) = ((b2) ⋅ (1 / 2)) + ((a2) ⋅ (1 / 2)) ((b2) + (a2)) ⋅ (1 / 2) = ((b2) ⋅ (1 / 2)) + ((a2) ⋅ (1 / 2))
3 (b2) ⋅ (1 / 2) = b if not (2 = 0), then (b2) ⋅ (1 / 2) = b
4 (a2) ⋅ (1 / 2) = a if not (2 = 0), then (a2) ⋅ (1 / 2) = a
5 ((b2) + (a2)) ⋅ (1 / 2) = b + ((a2) ⋅ (1 / 2)) if ((b2) + (a2)) ⋅ (1 / 2) = ((b2) ⋅ (1 / 2)) + ((a2) ⋅ (1 / 2)) and (b2) ⋅ (1 / 2) = b, then ((b2) + (a2)) ⋅ (1 / 2) = b + ((a2) ⋅ (1 / 2))
6 ((b2) + (a2)) ⋅ (1 / 2) = b + a if ((b2) + (a2)) ⋅ (1 / 2) = b + ((a2) ⋅ (1 / 2)) and (a2) ⋅ (1 / 2) = a, then ((b2) + (a2)) ⋅ (1 / 2) = b + a
7 ((b2) + (a2)) / 2 = b + a if ((b2) + (a2)) / 2 = ((b2) + (a2)) ⋅ (1 / 2) and ((b2) + (a2)) ⋅ (1 / 2) = b + a, then ((b2) + (a2)) / 2 = b + a
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