Proof: Doubling a Number

Let's prove the following theorem:

In step 2, we claim that:

a + a = (a ⋅ 1) + (a ⋅ 1)

And in step 3, we claim that:

(a ⋅ 1) + (a ⋅ 1) = a ⋅ (1 + 1)

Proof:

Proof Table

#	Claim	Reason
1	a = a ⋅ 1	a = a ⋅ 1 (multiplicative_identity_2)
2	a + a = (a ⋅ 1) + (a ⋅ 1)	if a = a ⋅ 1 and a = a ⋅ 1, then a + a = (a ⋅ 1) + (a ⋅ 1) (Substitution)
3	(a ⋅ 1) + (a ⋅ 1) = a ⋅ (1 + 1)	(a ⋅ 1) + (a ⋅ 1) = a ⋅ (1 + 1) (Distributive Property Variation)
4	1 + 1 = 2	1 + 1 = 2 (Fixed Value Statement)
5	a ⋅ (1 + 1) = a ⋅ 2	if 1 + 1 = 2, then a ⋅ (1 + 1) = a ⋅ 2 (Multiplicative Property of Equality Variation 1)
6	(a ⋅ 1) + (a ⋅ 1) = a ⋅ 2	if a ⋅ (1 + 1) = a ⋅ 2 and (a ⋅ 1) + (a ⋅ 1) = a ⋅ (1 + 1), then (a ⋅ 1) + (a ⋅ 1) = a ⋅ 2 (Transitive Property of Equality)
7	a + a = a ⋅ 2	if (a ⋅ 1) + (a ⋅ 1) = a ⋅ 2 and a + a = (a ⋅ 1) + (a ⋅ 1), then a + a = a ⋅ 2 (Transitive Property of Equality)

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