Proof: Example 2

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	2 > 0	2 > 0 (Fixed Value Statement)
2	(b ⋅ 2) - 0 = b ⋅ 2	(b ⋅ 2) - 0 = b ⋅ 2 (subtract_zero_example)
3	b ⋅ 2 = (b ⋅ 2) - 0	if (b ⋅ 2) - 0 = b ⋅ 2, then b ⋅ 2 = (b ⋅ 2) - 0 (Symmetric Property of Equality)
4	b ⋅ 2 > 0	if b > 0 and 2 > 0, then b ⋅ 2 > 0 (Greater Than Property 2)
5	(b ⋅ 2) - 0 > 0	if b ⋅ 2 > 0 and b ⋅ 2 = (b ⋅ 2) - 0, then (b ⋅ 2) - 0 > 0 (Transitive Property of Inequality 4)
6	not ((b ⋅ 2) - 0 = 0)	if (b ⋅ 2) - 0 > 0, then not ((b ⋅ 2) - 0 = 0) (Greater Than is Not Equal)
7	0 / ((b ⋅ 2) - 0) = 0	if not ((b ⋅ 2) - 0 = 0), then 0 / ((b ⋅ 2) - 0) = 0 (Zero Numerator Property)
8	0 - 0 = 0	0 - 0 = 0 (subtract_to_zero)
9	(0 - 0) / ((b ⋅ 2) - 0) = 0 / ((b ⋅ 2) - 0)	if 0 - 0 = 0, then (0 - 0) / ((b ⋅ 2) - 0) = 0 / ((b ⋅ 2) - 0) (Divide Both Sides)
10	(0 - 0) / ((b ⋅ 2) - 0) = 0	if (0 - 0) / ((b ⋅ 2) - 0) = 0 / ((b ⋅ 2) - 0) and 0 / ((b ⋅ 2) - 0) = 0, then (0 - 0) / ((b ⋅ 2) - 0) = 0 (Transitive Property of Equality)

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