Solutions to Problem Set 5

Math 310-01/02

9-15-2017

1. Premises: $\displaystyle
   \left\{\matrix{Q \land R \cr P \ifthen \lnot Q \cr (\lnot P \land R)
   \ifthen S \cr}\right.$

Prove: S.

$$\matrix{ \hfill 1. & Q \land R \hfill & \hbox{Premise} \hfill \cr \hfill 2. & P \ifthen \lnot Q \hfill & \hbox{Premise} \hfill \cr \hfill 3. & (\lnot P \land R) \ifthen S \hfill & \hbox{Premise} \hfill \cr \hfill 4. & Q \hfill & \hbox{Decomposing a conjunction (1)} \hfill \cr \hfill 5. & R \hfill & \hbox{Decomposing a conjunction (1)} \hfill \cr \hfill 6. & \lnot P \hfill & \hbox{Modus tollens (2,4)} \hfill \cr \hfill 7. & \lnot P \land R \hfill & \hbox{Constructing a conjunction (5,6)} \hfill \cr \hfill 8. & S \hfill & \hbox{Modus ponens (3,7)} \quad\halmos \hfill \cr}$$


2. Prove that if $x \in \real$ , then

$$x^2 - 6 x + 2 \cos x + 11 \ge 0.$$

Since squares are nonnegative,

$$(x - 3)^2 \ge 0.$$

Since $\cos x \ge -1$ ,

$$2 \cos x \ge -2.$$

Adding the two inequalities, I have

$$\eqalign{ (x - 3)^2 + 2 \cos x & \ge -2 \cr x^2 - 6 x + 9 + 2 \cos x & \ge -2 \cr x^2 - 6 x + 2 \cos x + 11 \ge 0 \cr} \quad\halmos$$


As he thinketh in his heart, so is he. - Proverbs 23:7


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