Proof: Distributive Applied

Let's prove the following theorem:

(a + b) ⋅ (b ⋅ (-1)) = ((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1))

Proof:

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Proof Table
# Claim Reason
1 (a + b) ⋅ (b ⋅ (-1)) = (a ⋅ (b ⋅ (-1))) + (b ⋅ (b ⋅ (-1))) (a + b) ⋅ (b ⋅ (-1)) = (a ⋅ (b ⋅ (-1))) + (b ⋅ (b ⋅ (-1)))
2 a ⋅ (b ⋅ (-1)) = (a ⋅ b) ⋅ (-1) a ⋅ (b ⋅ (-1)) = (a ⋅ b) ⋅ (-1)
3 b ⋅ (b ⋅ (-1)) = (b ⋅ b) ⋅ (-1) b ⋅ (b ⋅ (-1)) = (b ⋅ b) ⋅ (-1)
4 (a ⋅ (b ⋅ (-1))) + (b ⋅ (b ⋅ (-1))) = ((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1)) if b ⋅ (b ⋅ (-1)) = (b ⋅ b) ⋅ (-1) and a ⋅ (b ⋅ (-1)) = (a ⋅ b) ⋅ (-1), then (a ⋅ (b ⋅ (-1))) + (b ⋅ (b ⋅ (-1))) = ((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1))
5 (a + b) ⋅ (b ⋅ (-1)) = ((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1)) if (a ⋅ (b ⋅ (-1))) + (b ⋅ (b ⋅ (-1))) = ((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1)) and (a + b) ⋅ (b ⋅ (-1)) = (a ⋅ (b ⋅ (-1))) + (b ⋅ (b ⋅ (-1))), then (a + b) ⋅ (b ⋅ (-1)) = ((a ⋅ b) ⋅ (-1)) + ((b ⋅ b) ⋅ (-1))

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