Proof: Square Product Example

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	s / 2 = s ⋅ (1 / 2)	s / 2 = s ⋅ (1 / 2) (Division is Multiplication by Inverse)
2	(s / 2) ⋅ (s / 2) = ((s ⋅ (1 / 2)) ⋅ s) ⋅ (1 / 2)	if s / 2 = s ⋅ (1 / 2), then (s / 2) ⋅ (s / 2) = ((s ⋅ (1 / 2)) ⋅ s) ⋅ (1 / 2) (Substitution 16)
3	((s ⋅ (1 / 2)) ⋅ s) ⋅ (1 / 2) = ((s ⋅ s) ⋅ (1 / 2)) ⋅ (1 / 2)	((s ⋅ (1 / 2)) ⋅ s) ⋅ (1 / 2) = ((s ⋅ s) ⋅ (1 / 2)) ⋅ (1 / 2) (Swap Terms 2 and 3)
4	((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)
5	(1 / 2) ⋅ (1 / 2) = 1 / 4	(1 / 2) ⋅ (1 / 2) = 1 / 4 (Fixed Value Statement)
6	(s ⋅ s) ⋅ (1 / 4) = (s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2))	if (1 / 2) ⋅ (1 / 2) = 1 / 4, then (s ⋅ s) ⋅ (1 / 4) = (s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2)) (Multiplicative Property of Equality Variation 2)
7	(s ⋅ s) ⋅ (1 / 4) = ((s ⋅ s) ⋅ (1 / 2)) ⋅ (1 / 2)	if (s ⋅ s) ⋅ (1 / 4) = (s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2)) and ((s ⋅ s) ⋅ (1 / 2)) ⋅ (1 / 2) = (s ⋅ s) ⋅ ((1 / 2) ⋅ (1 / 2)), then (s ⋅ s) ⋅ (1 / 4) = ((s ⋅ s) ⋅ (1 / 2)) ⋅ (1 / 2) (Transitive Property of Equality Variation 1)
8	(s ⋅ s) ⋅ (1 / 4) = ((s ⋅ (1 / 2)) ⋅ s) ⋅ (1 / 2)	if (s ⋅ s) ⋅ (1 / 4) = ((s ⋅ s) ⋅ (1 / 2)) ⋅ (1 / 2) and ((s ⋅ (1 / 2)) ⋅ s) ⋅ (1 / 2) = ((s ⋅ s) ⋅ (1 / 2)) ⋅ (1 / 2), then (s ⋅ s) ⋅ (1 / 4) = ((s ⋅ (1 / 2)) ⋅ s) ⋅ (1 / 2) (Transitive Property of Equality Variation 1)
9	(s ⋅ s) ⋅ (1 / 4) = (s / 2) ⋅ (s / 2)	if (s ⋅ s) ⋅ (1 / 4) = ((s ⋅ (1 / 2)) ⋅ s) ⋅ (1 / 2) and (s / 2) ⋅ (s / 2) = ((s ⋅ (1 / 2)) ⋅ s) ⋅ (1 / 2), then (s ⋅ s) ⋅ (1 / 4) = (s / 2) ⋅ (s / 2) (Transitive Property of Equality Variation 1)

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