Proof: Sum Squared Theorem

Let's prove the following theorem:

(a + b) ⋅ (a + b) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b)

a b a b

Proof:

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Proof Table
# Claim Reason
1 (a + b) ⋅ (a + b) = ((a + b) ⋅ a) + ((a + b) ⋅ b) (a + b) ⋅ (a + b) = ((a + b) ⋅ a) + ((a + b) ⋅ b)
2 (a + b) ⋅ a = (a ⋅ a) + (a ⋅ b) (a + b) ⋅ a = (a ⋅ a) + (a ⋅ b)
3 (a + b) ⋅ b = (a ⋅ b) + (b ⋅ b) (a + b) ⋅ b = (a ⋅ b) + (b ⋅ b)
4 ((a + b) ⋅ a) + ((a + b) ⋅ b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) if (a + b) ⋅ a = (a ⋅ a) + (a ⋅ b) and (a + b) ⋅ b = (a ⋅ b) + (b ⋅ b), then ((a + b) ⋅ a) + ((a + b) ⋅ b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b))
5 (a + b) ⋅ (a + b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) if (a + b) ⋅ (a + b) = ((a + b) ⋅ a) + ((a + b) ⋅ b) and ((a + b) ⋅ a) + ((a + b) ⋅ b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)), then (a + b) ⋅ (a + b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b))
6 (((a ⋅ a) + (a ⋅ b)) + (a ⋅ b)) + (b ⋅ b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) (((a ⋅ a) + (a ⋅ b)) + (a ⋅ b)) + (b ⋅ b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b))
7 (a ⋅ b) + (a ⋅ b) = (a ⋅ b) ⋅ 2 (a ⋅ b) + (a ⋅ b) = (a ⋅ b) ⋅ 2
8 (((a ⋅ a) + (a ⋅ b)) + (a ⋅ b)) + (b ⋅ b) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b) if (a ⋅ b) + (a ⋅ b) = (a ⋅ b) ⋅ 2, then (((a ⋅ a) + (a ⋅ b)) + (a ⋅ b)) + (b ⋅ b) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b)
9 ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b) if (((a ⋅ a) + (a ⋅ b)) + (a ⋅ b)) + (b ⋅ b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) and (((a ⋅ a) + (a ⋅ b)) + (a ⋅ b)) + (b ⋅ b) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b), then ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b)
10 (a + b) ⋅ (a + b) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b) if (a + b) ⋅ (a + b) = ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) and ((a ⋅ a) + (a ⋅ b)) + ((a ⋅ b) + (b ⋅ b)) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b), then (a + b) ⋅ (a + b) = ((a ⋅ a) + ((a ⋅ b) ⋅ 2)) + (b ⋅ b)

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