Proof: Inequality Problem 4 2

Let's prove the following theorem:

if (x + 10) ⋅ (1 / (-5)) > 3, then x < -25

Proof:

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Given

1	(x + 10) ⋅ (1 / (-5)) > 3

Proof Table

#	Claim	Reason
1	-5 < 0	-5 < 0 (Fixed Value Statement)
2	((x + 10) ⋅ (1 / (-5))) ⋅ (-5) < 3 ⋅ (-5)	if -5 < 0 and (x + 10) ⋅ (1 / (-5)) > 3, then ((x + 10) ⋅ (1 / (-5))) ⋅ (-5) < 3 ⋅ (-5) (Inequality Change)
3	3 ⋅ (-5) = -15	3 ⋅ (-5) = -15 (Fixed Value Statement)
4	(1 / (-5)) ⋅ (-5) = 1	(1 / (-5)) ⋅ (-5) = 1 (Fixed Value Statement)
5	((x + 10) ⋅ (1 / (-5))) ⋅ (-5) = (x + 10) ⋅ 1	if (1 / (-5)) ⋅ (-5) = 1, then ((x + 10) ⋅ (1 / (-5))) ⋅ (-5) = (x + 10) ⋅ 1 (Apply Associative Multiply)
6	(x + 10) ⋅ 1 < -15	if 3 ⋅ (-5) = -15 and ((x + 10) ⋅ (1 / (-5))) ⋅ (-5) = (x + 10) ⋅ 1 and ((x + 10) ⋅ (1 / (-5))) ⋅ (-5) < 3 ⋅ (-5), then (x + 10) ⋅ 1 < -15 (Double Transitive Property Less)
7	x + 10 < -15	if (x + 10) ⋅ 1 < -15, then x + 10 < -15 (Multiply by 1 Less)
8	(x + 10) + (-10) < (-15) + (-10)	if x + 10 < -15, then (x + 10) + (-10) < (-15) + (-10) (Inequality Addition Property)
9	(-15) + (-10) = -25	(-15) + (-10) = -25 (Fixed Value Statement)
10	10 + (-10) = 0	10 + (-10) = 0 (Fixed Value Statement)
11	(x + 10) + (-10) = x + 0	if 10 + (-10) = 0, then (x + 10) + (-10) = x + 0 (Substitution 16 Pre)
12	x + 0 < -25	if (-15) + (-10) = -25 and (x + 10) + (-10) = x + 0 and (x + 10) + (-10) < (-15) + (-10), then x + 0 < -25 (Double Transitive Property Less)
13	x < -25	if x + 0 < -25, then x < -25 (Add by 0 Less)

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