Proof: Divide Simplify 2

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	(b ⋅ d) ⋅ c = (b ⋅ c) ⋅ d	(b ⋅ d) ⋅ c = (b ⋅ c) ⋅ d (Swap 2 and 3 Theorem)
2	((b ⋅ d) ⋅ c) / d = ((b ⋅ c) ⋅ d) / d	if (b ⋅ d) ⋅ c = (b ⋅ c) ⋅ d, then ((b ⋅ d) ⋅ c) / d = ((b ⋅ c) ⋅ d) / d (Divide Both Sides)
3	d / d = 1	d / d = 1 (Multiplicative Inverse Property)
4	((b ⋅ c) ⋅ d) / d = (b ⋅ c) ⋅ (d / d)	((b ⋅ c) ⋅ d) / d = (b ⋅ c) ⋅ (d / d) (associative_property)
5	((b ⋅ c) ⋅ d) / d = (b ⋅ c) ⋅ 1	if ((b ⋅ c) ⋅ d) / d = (b ⋅ c) ⋅ (d / d) and d / d = 1, then ((b ⋅ c) ⋅ d) / d = (b ⋅ c) ⋅ 1 (Substitution 12)
6	(b ⋅ c) ⋅ 1 = b ⋅ c	(b ⋅ c) ⋅ 1 = b ⋅ c (Multiplicative Identity)
7	((b ⋅ c) ⋅ d) / d = b ⋅ c	if ((b ⋅ c) ⋅ d) / d = (b ⋅ c) ⋅ 1 and (b ⋅ c) ⋅ 1 = b ⋅ c, then ((b ⋅ c) ⋅ d) / d = b ⋅ c (Transitive Property of Equality)
8	((b ⋅ d) ⋅ c) / d = b ⋅ c	if ((b ⋅ d) ⋅ c) / d = ((b ⋅ c) ⋅ d) / d and ((b ⋅ c) ⋅ d) / d = b ⋅ c, then ((b ⋅ d) ⋅ c) / d = b ⋅ c (Transitive Property of Equality)
9	((b ⋅ d) ⋅ c) / d = (b ⋅ d) ⋅ (c / d)	((b ⋅ d) ⋅ c) / d = (b ⋅ d) ⋅ (c / d) (associative_property)
10	(b ⋅ d) ⋅ (c / d) = b ⋅ c	if ((b ⋅ d) ⋅ c) / d = (b ⋅ d) ⋅ (c / d) and ((b ⋅ d) ⋅ c) / d = b ⋅ c, then (b ⋅ d) ⋅ (c / d) = b ⋅ c (Transitive Property of Equality Variation 2)

Comments

Please log in to add comments