Proof: Square Product Example

Let's prove the following theorem:

We know that:

(1 / 2) ⋅ (1 / 2) = 1 / 4

Thus we claim that:

(s ⋅ s) ⋅ (1 / 4) = (s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2))

Finally, we show that:

(s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2)) = (s / 2) ⋅ (s / 2)

Proof:

Proof Table

#	Claim	Reason
1	(1 / 2) ⋅ (1 / 2) = 1 / 4	(1 / 2) ⋅ (1 / 2) = 1 / 4 (Fixed Value Statement)
2	(s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2)) = (s ⋅ s) ⋅ (1 / 4)	if (1 / 2) ⋅ (1 / 2) = 1 / 4, then (s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2)) = (s ⋅ s) ⋅ (1 / 4) (Substitution)
3	(s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2)) = ((s ⋅ s) ⋅ (1 / 2)) ⋅ (1 / 2)	(s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2)) = ((s ⋅ s) ⋅ (1 / 2)) ⋅ (1 / 2) (associative_property_of_multiplication_2)
4	((s ⋅ s) ⋅ (1 / 2)) ⋅ (1 / 2) = (s / 2) ⋅ (s / 2)	((s ⋅ s) ⋅ (1 / 2)) ⋅ (1 / 2) = (s / 2) ⋅ (s / 2) (inverse_example)
5	(s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2)) = (s / 2) ⋅ (s / 2)	if (s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2)) = ((s ⋅ s) ⋅ (1 / 2)) ⋅ (1 / 2) and ((s ⋅ s) ⋅ (1 / 2)) ⋅ (1 / 2) = (s / 2) ⋅ (s / 2), then (s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2)) = (s / 2) ⋅ (s / 2) (Transitive Property of Equality)
6	(s ⋅ s) ⋅ (1 / 4) = (s / 2) ⋅ (s / 2)	if (s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2)) = (s ⋅ s) ⋅ (1 / 4) and (s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2)) = (s / 2) ⋅ (s / 2), then (s ⋅ s) ⋅ (1 / 4) = (s / 2) ⋅ (s / 2) (Transitive Property of Equality Variation 2)

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