Proof: Distributive Property Variation 3

Let's prove the following theorem:

(a + b) ⋅ c = (a ⋅ c) + (b ⋅ c)

Proof:

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

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