Proof: Distribute Subtract

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	a - b = a + (b ⋅ (-1))	a - b = a + (b ⋅ (-1)) (Subtraction Property)
2	(a - b) ⋅ c = (a + (b ⋅ (-1))) ⋅ c	if a - b = a + (b ⋅ (-1)), then (a - b) ⋅ c = (a + (b ⋅ (-1))) ⋅ c (Multiplicative Property of Equality)
3	(a + (b ⋅ (-1))) ⋅ c = (a ⋅ c) + ((b ⋅ (-1)) ⋅ c)	(a + (b ⋅ (-1))) ⋅ c = (a ⋅ c) + ((b ⋅ (-1)) ⋅ c) (Distributive Property Variation 3)
4	(b ⋅ (-1)) ⋅ c = (b ⋅ c) ⋅ (-1)	(b ⋅ (-1)) ⋅ c = (b ⋅ c) ⋅ (-1) (Swap Terms 2 and 3)
5	(a ⋅ c) + ((b ⋅ (-1)) ⋅ c) = (a ⋅ c) + ((b ⋅ c) ⋅ (-1))	if (b ⋅ (-1)) ⋅ c = (b ⋅ c) ⋅ (-1), then (a ⋅ c) + ((b ⋅ (-1)) ⋅ c) = (a ⋅ c) + ((b ⋅ c) ⋅ (-1)) (Substitution)
6	(a ⋅ c) + ((b ⋅ c) ⋅ (-1)) = (a ⋅ c) - (b ⋅ c)	(a ⋅ c) + ((b ⋅ c) ⋅ (-1)) = (a ⋅ c) - (b ⋅ c) (subtraction_example)
7	(a ⋅ c) + ((b ⋅ (-1)) ⋅ c) = (a ⋅ c) - (b ⋅ c)	if (a ⋅ c) + ((b ⋅ c) ⋅ (-1)) = (a ⋅ c) - (b ⋅ c) and (a ⋅ c) + ((b ⋅ (-1)) ⋅ c) = (a ⋅ c) + ((b ⋅ c) ⋅ (-1)), then (a ⋅ c) + ((b ⋅ (-1)) ⋅ c) = (a ⋅ c) - (b ⋅ c) (Transitive Property of Equality)
8	(a + (b ⋅ (-1))) ⋅ c = (a ⋅ c) - (b ⋅ c)	if (a ⋅ c) + ((b ⋅ (-1)) ⋅ c) = (a ⋅ c) - (b ⋅ c) and (a + (b ⋅ (-1))) ⋅ c = (a ⋅ c) + ((b ⋅ (-1)) ⋅ c), then (a + (b ⋅ (-1))) ⋅ c = (a ⋅ c) - (b ⋅ c) (Transitive Property of Equality)
9	(a - b) ⋅ c = (a ⋅ c) - (b ⋅ c)	if (a + (b ⋅ (-1))) ⋅ c = (a ⋅ c) - (b ⋅ c) and (a - b) ⋅ c = (a + (b ⋅ (-1))) ⋅ c, then (a - b) ⋅ c = (a ⋅ c) - (b ⋅ c) (Transitive Property of Equality)

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