Solutions to Problem Set 6

Math 310-01/02

9-15-2017

1. Premises: $\displaystyle
   \left\{\matrix{A \land \lnot B \cr (A \lor B) \ifthen \lnot C
   \cr}\right.$

Prove: $\lnot(C \lor B)$ .

$$\matrix{ \hfill 1. & A \land \lnot B \hfill & \hbox{Premise} \hfill \cr \hfill 2. & (A \lor B) \ifthen \lnot C \hfill & \hbox{Premise} \hfill \cr \hfill 3. & A \hfill & \hbox{Decomposing a conjunction (1)} \hfill \cr \hfill 4. & \lnot B \hfill & \hbox{Decomposing a conjunction (1)} \hfill \cr \hfill 5. & A \lor B \hfill & \hbox{Constructing a disjunction (3)} \hfill \cr \hfill 6. & \lnot C \hfill & \hbox{Modus ponens (2,5)} \hfill \cr \hfill 7. & \lnot B \land \lnot C \hfill & \hbox{Constructing a conjunction (4,6)} \hfill \cr \hfill 8. & \lnot (B \lor C) \hfill & \hbox{DeMorgan (7)} \quad\halmos \hfill \cr}$$


2. Use the triangle inequality $|x| + |y| \ge |x + y|$ to prove:

(a) If $x \in \real$ , then $|x + 9| + |x - 5| \ge 14$ .

(b) If $a, b, c \in \real$ , then $|a - b| + |c - b| \ge |a - c|$ .

(a) Note that $|x - 5| = |5 - x|$ . So

$$\eqalign{ |x + 9| + |x - 5| & = |x + 9| + |5 - x| \cr & \ge |(x + 9) + (5 - x)| \cr & = |14| \cr & = 14 \cr} \quad\halmos$$

(b) Note that $|c - b| = |b -
   c|$ . So

$$\eqalign{ |a - b| + |c - b| & = |a - b| + |b - c| \cr & \ge |(a - b) + (b - c)| \cr & = |a - c| \cr} \quad\halmos$$


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