1.      (3 pts)Use symbols (p, q, r) as needed to write the logical form of the argument. Either use truth tables to determine the validity of the argument or identify the rule of inference/error that guarantees its validity/falsity.

Argument

Symbolic Representation (2 pts for either)

If at least one of these two numbers is divisible by 6, then the product of these two numbers is divisible by 6.


The product of these two numbers is not divisible by 6.

\ Neither of these two numbers is divisible by 6.

p ® q


~q

\~p

p Ú q ® r

 

 

~r

 

\~(p Ú q)

Rule of inference: Valid by Modus Tollens. (1 pt)
Alternate solutions: (or 1 pt)

p q

p®q

~q

critical rows

~p

T T

T

F

 

 

T F

F

T

 

 

F T

T

F

 

 

F F

T

T

y valid

T

 

p q r

p Ú q ® r

~r

crit rows

~(p Ú q)

T T T

T

F

 

 

T T F

F

T

 

 

T F T

T

F

 

 

T F F

F

T

 

 

F T T

T

F

 

 

F T F

F

T

 

 

F F T

T

F

 

 

F F F

T

T

y valid

T

 

p q r

(p Ú q ® r) Ù ~r ® ~(p Ú q)

T T T

T

T T F

T

T F T

T

T F F

T

F T T

T

F T F

T

F F T

T

F F F

T

 

A tautology so valid

 

2.      (4 pts)Design a circuit for this input/output table. Use either Theorem 1.1.1 or a Karnaugh map to reduce this circuit. Then draw your reduced circuit.

 

p

q

r

S

0

0

0

1

0

0

1

0

0

1

0

1

0

1

1

1

1

0

0

0

1

0

1

0

1

1

0

0

1

1

1

0

p’q’r’ + p’qr’ + p’qr º 
p’(q’r’ + qr’ + qr) º
p’(q’r’ + qr’ + qr’ + qr) º
p’(r’(q’ + q) + q(r’ + r)) º
p’(r’t + qt) º
p’r’ + p’q   (2pts)

Karnaugh Map: (or 1pt)

p\qr

00

01

11

10

0

1

 

1

1

1

 

 

 

 

S = p’r’ + p’q (or 2pts)

disjunctive normal form (1pt)
distributive law
idempotent law
distributive law
negation law
identity law and distributive law

(1 pt also credit valid disjunctive normal form dwg.)

3.      (3 pts)Use Theorem 1.1.1 to show that ~(p Ú q) Ú ~p º ~p. (Show your work, you do not have to provide the name for each law you utilize, but write it down if you know it.)
 

~(p Ú q) Ú ~p

º

(~p Ù ~q) Ú ~p

by Demorgan’s Law

(2pts)

 

º

~p Ú (~p Ù ~q)

by assoc. and comm..

 

 

º

~p

by absorption

(1pt)