Proof: Simplify 3

Let's prove the following theorem:

if not (2 = 0), then ((b ⋅ 2) + (a ⋅ 2)) / 2 = b + a

Proof:

View as a tree | View dependent proofs | Try proving it

Given

1	not (2 = 0)

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	if not (2 = 0), then (b ⋅ 2) ⋅ (1 / 2) = b (Simplify 2)
4	(a ⋅ 2) ⋅ (1 / 2) = a	if not (2 = 0), then (a ⋅ 2) ⋅ (1 / 2) = a (Simplify 2)
5	((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = b + ((a ⋅ 2) ⋅ (1 / 2))	if (b ⋅ 2) ⋅ (1 / 2) = b and ((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = ((b ⋅ 2) ⋅ (1 / 2)) + ((a ⋅ 2) ⋅ (1 / 2)), then ((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = b + ((a ⋅ 2) ⋅ (1 / 2)) (Substitution 6)
6	((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = b + a	if (a ⋅ 2) ⋅ (1 / 2) = a and ((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = b + ((a ⋅ 2) ⋅ (1 / 2)), then ((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = b + a (Substitution 2)
7	((b ⋅ 2) + (a ⋅ 2)) / 2 = b + a	if ((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2) = b + a and ((b ⋅ 2) + (a ⋅ 2)) / 2 = ((b ⋅ 2) + (a ⋅ 2)) ⋅ (1 / 2), then ((b ⋅ 2) + (a ⋅ 2)) / 2 = b + a (Transitive Property of Equality)

Comments

Please log in to add comments