Proof: Simplify 3

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	((b ⋅ 2) + (a ⋅ 2)) / 2 = ((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2)	((b ⋅ 2) + (a ⋅ 2)) / 2 = ((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) (Division is Multiplication by Inverse)
2	((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = ((b ⋅ 2) ⋅ (1 / 2)) + ((a ⋅ 2) ⋅ (1 / 2))	((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = ((b ⋅ 2) ⋅ (1 / 2)) + ((a ⋅ 2) ⋅ (1 / 2)) (Distributive Property Variation 3)
3	(b ⋅ 2) ⋅ (1 / 2) = b	(b ⋅ 2) ⋅ (1 / 2) = b (simplify_2)
4	(a ⋅ 2) ⋅ (1 / 2) = a	(a ⋅ 2) ⋅ (1 / 2) = a (simplify_2)
5	((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = b + ((a ⋅ 2) ⋅ (1 / 2))	if ((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = ((b ⋅ 2) ⋅ (1 / 2)) + ((a ⋅ 2) ⋅ (1 / 2)) and (b ⋅ 2) ⋅ (1 / 2) = b, then ((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = b + ((a ⋅ 2) ⋅ (1 / 2)) (Term 1 Substitution)
6	((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = b + a	if ((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = b + ((a ⋅ 2) ⋅ (1 / 2)) and (a ⋅ 2) ⋅ (1 / 2) = a, then ((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = b + a (Term 2 Substitution)
7	((b ⋅ 2) + (a ⋅ 2)) / 2 = b + a	if ((b ⋅ 2) + (a ⋅ 2)) / 2 = ((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) and ((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = b + a, then ((b ⋅ 2) + (a ⋅ 2)) / 2 = b + a (Transitive Property of Equality)

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