Proof: Algebra 10 Help

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	(b ⋅ (-1)) / c = (b ⋅ (-1)) ⋅ (1 / c)	(b ⋅ (-1)) / c = (b ⋅ (-1)) ⋅ (1 / c) (Division is Multiplication by Inverse)
2	(b ⋅ (-1)) ⋅ (1 / c) = (b ⋅ (1 / c)) ⋅ (-1)	(b ⋅ (-1)) ⋅ (1 / c) = (b ⋅ (1 / c)) ⋅ (-1) (Swap 2 and 3 Theorem)
3	b ⋅ (1 / c) = b / c	b ⋅ (1 / c) = b / c (division_theorem)
4	(b ⋅ (1 / c)) ⋅ (-1) = (b / c) ⋅ (-1)	if b ⋅ (1 / c) = b / c, then (b ⋅ (1 / c)) ⋅ (-1) = (b / c) ⋅ (-1) (Multiplicative Property of Equality)
5	(b ⋅ (-1)) ⋅ (1 / c) = (b / c) ⋅ (-1)	if (b ⋅ (-1)) ⋅ (1 / c) = (b ⋅ (1 / c)) ⋅ (-1) and (b ⋅ (1 / c)) ⋅ (-1) = (b / c) ⋅ (-1), then (b ⋅ (-1)) ⋅ (1 / c) = (b / c) ⋅ (-1) (Transitive Property of Equality)
6	(b ⋅ (-1)) / c = (b / c) ⋅ (-1)	if (b ⋅ (-1)) / c = (b ⋅ (-1)) ⋅ (1 / c) and (b ⋅ (-1)) ⋅ (1 / c) = (b / c) ⋅ (-1), then (b ⋅ (-1)) / c = (b / c) ⋅ (-1) (Transitive Property of Equality)
7	(b ⋅ (-1)) / c = b ⋅ ((-1) / c)	(b ⋅ (-1)) / c = b ⋅ ((-1) / c) (associative_property)
8	b ⋅ ((-1) / c) = (b / c) ⋅ (-1)	if (b ⋅ (-1)) / c = b ⋅ ((-1) / c) and (b ⋅ (-1)) / c = (b / c) ⋅ (-1), then b ⋅ ((-1) / c) = (b / c) ⋅ (-1) (Transitive Property of Equality Variation 2)
9	a + (b ⋅ ((-1) / c)) = a + ((b / c) ⋅ (-1))	if b ⋅ ((-1) / c) = (b / c) ⋅ (-1), then a + (b ⋅ ((-1) / c)) = a + ((b / c) ⋅ (-1)) (Substitution)
10	a + ((b / c) ⋅ (-1)) = a - (b / c)	a + ((b / c) ⋅ (-1)) = a - (b / c) (subtraction_example)
11	a + (b ⋅ ((-1) / c)) = a - (b / c)	if a + (b ⋅ ((-1) / c)) = a + ((b / c) ⋅ (-1)) and a + ((b / c) ⋅ (-1)) = a - (b / c), then a + (b ⋅ ((-1) / c)) = a - (b / c) (Transitive Property of Equality)

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