Proof: Manipulation 2

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	(a ⋅ b) ⋅ (-1) = a ⋅ (b ⋅ (-1))	(a ⋅ b) ⋅ (-1) = a ⋅ (b ⋅ (-1)) (Associative Property of Multiplication)
2	((a ⋅ b) ⋅ (-1)) ⋅ 2 = (a ⋅ (b ⋅ (-1))) ⋅ 2	if (a ⋅ b) ⋅ (-1) = a ⋅ (b ⋅ (-1)), then ((a ⋅ b) ⋅ (-1)) ⋅ 2 = (a ⋅ (b ⋅ (-1))) ⋅ 2 (Multiplicative Property of Equality)
3	((a ⋅ b) ⋅ (-1)) ⋅ 2 = (a ⋅ b) ⋅ ((-1) ⋅ 2)	((a ⋅ b) ⋅ (-1)) ⋅ 2 = (a ⋅ b) ⋅ ((-1) ⋅ 2) (Associative Property of Multiplication)
4	(-1) ⋅ 2 = -2	(-1) ⋅ 2 = -2 (Fixed Value Statement)
5	(a ⋅ b) ⋅ ((-1) ⋅ 2) = (a ⋅ b) ⋅ (-2)	if (-1) ⋅ 2 = -2, then (a ⋅ b) ⋅ ((-1) ⋅ 2) = (a ⋅ b) ⋅ (-2) (Multiplicative Property of Equality Variation 1)
6	((a ⋅ b) ⋅ (-1)) ⋅ 2 = (a ⋅ b) ⋅ (-2)	if ((a ⋅ b) ⋅ (-1)) ⋅ 2 = (a ⋅ b) ⋅ ((-1) ⋅ 2) and (a ⋅ b) ⋅ ((-1) ⋅ 2) = (a ⋅ b) ⋅ (-2), then ((a ⋅ b) ⋅ (-1)) ⋅ 2 = (a ⋅ b) ⋅ (-2) (Transitive Property of Equality)
7	(a ⋅ (b ⋅ (-1))) ⋅ 2 = (a ⋅ b) ⋅ (-2)	if ((a ⋅ b) ⋅ (-1)) ⋅ 2 = (a ⋅ (b ⋅ (-1))) ⋅ 2 and ((a ⋅ b) ⋅ (-1)) ⋅ 2 = (a ⋅ b) ⋅ (-2), then (a ⋅ (b ⋅ (-1))) ⋅ 2 = (a ⋅ b) ⋅ (-2) (Transitive Property of Equality Variation 2)

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