Proof: Subtract Reorder

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	(a + b) - c = (a + b) + (c ⋅ (-1))	(a + b) - c = (a + b) + (c ⋅ (-1)) (Subtraction Property)
2	(a + b) + (c ⋅ (-1)) = a + (b + (c ⋅ (-1)))	(a + b) + (c ⋅ (-1)) = a + (b + (c ⋅ (-1))) (Associative Property of Addition)
3	b + (c ⋅ (-1)) = (c ⋅ (-1)) + b	b + (c ⋅ (-1)) = (c ⋅ (-1)) + b (Commutative Property of Addition)
4	a + (b + (c ⋅ (-1))) = a + ((c ⋅ (-1)) + b)	if b + (c ⋅ (-1)) = (c ⋅ (-1)) + b, then a + (b + (c ⋅ (-1))) = a + ((c ⋅ (-1)) + b) (Substitution)
5	(a + b) + (c ⋅ (-1)) = a + ((c ⋅ (-1)) + b)	if a + (b + (c ⋅ (-1))) = a + ((c ⋅ (-1)) + b) and (a + b) + (c ⋅ (-1)) = a + (b + (c ⋅ (-1))), then (a + b) + (c ⋅ (-1)) = a + ((c ⋅ (-1)) + b) (Transitive Property of Equality)
6	(a + b) - c = a + ((c ⋅ (-1)) + b)	if (a + b) + (c ⋅ (-1)) = a + ((c ⋅ (-1)) + b) and (a + b) - c = (a + b) + (c ⋅ (-1)), then (a + b) - c = a + ((c ⋅ (-1)) + b) (Transitive Property of Equality)
7	a + ((c ⋅ (-1)) + b) = (a + (c ⋅ (-1))) + b	a + ((c ⋅ (-1)) + b) = (a + (c ⋅ (-1))) + b (associative)
8	a + (c ⋅ (-1)) = a - c	a + (c ⋅ (-1)) = a - c (subtraction_example)
9	(a + (c ⋅ (-1))) + b = (a - c) + b	if a + (c ⋅ (-1)) = a - c, then (a + (c ⋅ (-1))) + b = (a - c) + b (Additive Property of Equality)
10	a + ((c ⋅ (-1)) + b) = (a - c) + b	if (a + (c ⋅ (-1))) + b = (a - c) + b and a + ((c ⋅ (-1)) + b) = (a + (c ⋅ (-1))) + b, then a + ((c ⋅ (-1)) + b) = (a - c) + b (Transitive Property of Equality)
11	(a + b) - c = (a - c) + b	if a + ((c ⋅ (-1)) + b) = (a - c) + b and (a + b) - c = a + ((c ⋅ (-1)) + b), then (a + b) - c = (a - c) + b (Transitive Property of Equality)

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