Proof: Distributive Property Variation 2
Let's prove the following theorem:
(b + c) ⋅ a = (a ⋅ b) + (a ⋅ c)
Proof:
# | Claim | Reason |
---|---|---|
1 | a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c) | a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c) |
2 | (b + c) ⋅ a = a ⋅ (b + c) | (b + c) ⋅ a = a ⋅ (b + c) |
3 | (b + c) ⋅ a = (a ⋅ b) + (a ⋅ c) | if (b + c) ⋅ a = a ⋅ (b + c) and a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c), then (b + c) ⋅ a = (a ⋅ b) + (a ⋅ c) |
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