Proof: Square Root Example 2

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	(3 / 4) ⋅ (s ⋅ s) = 3 ⋅ ((s ⋅ s) ⋅ (1 / 4))	(3 / 4) ⋅ (s ⋅ s) = 3 ⋅ ((s ⋅ s) ⋅ (1 / 4)) (Shuffle Example)
2	square root of ((3 / 4) ⋅ (s ⋅ s)) = square root of (3 ⋅ ((s ⋅ s) ⋅ (1 / 4)))	if (3 / 4) ⋅ (s ⋅ s) = 3 ⋅ ((s ⋅ s) ⋅ (1 / 4)), then square root of ((3 / 4) ⋅ (s ⋅ s)) = square root of (3 ⋅ ((s ⋅ s) ⋅ (1 / 4))) (Substitution)
3	square root of (3 ⋅ ((s ⋅ s) ⋅ (1 / 4))) = (square root of 3) ⋅ (square root of ((s ⋅ s) ⋅ (1 / 4)))	square root of (3 ⋅ ((s ⋅ s) ⋅ (1 / 4))) = (square root of 3) ⋅ (square root of ((s ⋅ s) ⋅ (1 / 4))) (Product of Square Roots)
4	square root of ((s ⋅ s) ⋅ (1 / 4)) = s / 2	square root of ((s ⋅ s) ⋅ (1 / 4)) = s / 2 (square_root_example)
5	(square root of 3) ⋅ (square root of ((s ⋅ s) ⋅ (1 / 4))) = (square root of 3) ⋅ (s / 2)	if square root of ((s ⋅ s) ⋅ (1 / 4)) = s / 2, then (square root of 3) ⋅ (square root of ((s ⋅ s) ⋅ (1 / 4))) = (square root of 3) ⋅ (s / 2) (Multiplicative Property of Equality Variation 1)
6	square root of (3 ⋅ ((s ⋅ s) ⋅ (1 / 4))) = (square root of 3) ⋅ (s / 2)	if square root of (3 ⋅ ((s ⋅ s) ⋅ (1 / 4))) = (square root of 3) ⋅ (square root of ((s ⋅ s) ⋅ (1 / 4))) and (square root of 3) ⋅ (square root of ((s ⋅ s) ⋅ (1 / 4))) = (square root of 3) ⋅ (s / 2), then square root of (3 ⋅ ((s ⋅ s) ⋅ (1 / 4))) = (square root of 3) ⋅ (s / 2) (Transitive Property of Equality)
7	square root of ((3 / 4) ⋅ (s ⋅ s)) = (square root of 3) ⋅ (s / 2)	if square root of ((3 / 4) ⋅ (s ⋅ s)) = square root of (3 ⋅ ((s ⋅ s) ⋅ (1 / 4))) and square root of (3 ⋅ ((s ⋅ s) ⋅ (1 / 4))) = (square root of 3) ⋅ (s / 2), then square root of ((3 / 4) ⋅ (s ⋅ s)) = (square root of 3) ⋅ (s / 2) (Transitive Property of Equality)

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