1. Propositional Logic
Solutions
Exercises A.
1. What order of precedence does DERIVE use for the five predefined logical operators?
Solution
Author and simplify various DERIVE expressions to observe how DERIVE drops certain
parentheses and how DERIVE evaluates these expressions. The following expressions are
useful in determining operator precedence.
TRUTH_TABLE(p, q, NOT(p) AND q, NOT(p AND q))
TRUTH_TABLE(p, q, r, (p AND q) OR r, p AND (q OR r), (p OR q) AND r, p OR (q
AND r))
TRUTH_TABLE(p, q, r, (p AND q) XOR r, p AND (q XOR r), (p XOR q) AND r, p
XOR (q AND r))
TRUTH_TABLE(p, q, r, (p XOR q) OR r, p XOR (q OR r), (p OR q) XOR r, p OR (q
XOR r))
TRUTH_TABLE(p, q, r, (p IMP q) OR r, p IMP (q OR r), (p OR q) IMP r, p OR (q IMP
r))
The logical operator NOT has the highest precedence. Next in order of precedence is the
operator AND. Then come OR and XOR at the same level with right to left evaluation
when they occur in the same expression without parentheses. The operator IMP has the
lowest precedence of the five predefined logical operators.
2. Define the biconditional function IFF of two variables. For example, IFF(true, false)
should simplify to false. Check your definition of IFF with a truth table.
Solution
Enter the following DERIVE expressions and simplify the second expression.
IFF(p, q) := (p IMP q) AND (q IMP p)
TRUTH_TABLE(p, q, IFF(p, q))
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3. Use DERIVE truth tables to determine whether each of the following is a contradiction, a
tautology, or neither. Explain.
~p  p
p  ~p
(p  q)  (q  p)
(p  ~p)  (p  ~p)
Solution
Observe the values in the truth tables generated by the following DERIVE expression.
TRUTH_TABLE(p, NOT(p) AND p, p IMP NOT(p), p AND NOT(p) IMP p OR
NOT(p))
The first propositional form in the DERIVE expression is a contradiction because it is
false in all cases. The second form is neither a contradiction nor a tautology because it
has a true case and a false case. The third form is a tautology because it is true in all
cases
The propositional form involving both p and q can be investigated by simplifying the
following DERIVE expression.
TRUTH_TABLE(p, q, IFF(p IMP q, q IMP p))
This propositional form is neither a contradiction nor a tautology because it has both true
and false values.
4. Verify that the operator < on the set {-2, -1, 0, 1, 2} is not commutative. Explain.
Hint: Simplify the expression:
VECTOR( VECTOR( IFF (x < y , y < x), x, -2, 2), y, -2, 2)
Solution
Since false cases are observed when the DERIVE expression:
VECTOR(VECTOR(IFF(x<y, y<x),x, -2, 2), y, -2, 2)
is simplified, the operator < is not commutative.
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Exercises B.
So that we can use common names for the functions that we define in DERIVE, we
will override one default choice for input of function names. To allow case-sensitive names,
use the DERIVE command:
Declare Algebra State Input
and choose Sensitive Case.
expression.
Another way to change to this state is to author the following
CaseMode := Sensitive
1. In many applications of logic, such as in digital circuit design, the integers 0 and 1
correspond to the logic values false and true, respectively. With this correspondence in
mind, we will develop a system of logic using these integers rather than logic values.
Use simple arithmetic operations to define numerical DERIVE functions not and and that
behave as their names indicate:
not(p) := 1 – p
and(p, q) := pq
For example, some values of these functions should be:
not(0) = 1
and(1, 1) = 1
and(0, 1) = 0.
Verify that both functions yield correct results in all cases.
Solution
Author the following DERIVE expressions and simplify the VECTOR and APPEND
expressions to show that the functions not and and correctly implement the corresponding
logical operators:
CaseMode := Sensitive
not(p) := 1 - p
VECTOR([p, not(p)], p, 0, 1)
and(p, q) := p*q
A := VECTOR(VECTOR([p, q, and (p, q)], q, 0, 1), p, 0, 1)
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APPEND(A SUB 1, A SUB 2)
2. Suppose that you are given only the definitions of the logic operators negation and
conjunction with truth tables. Define the logic operators disjunction, conditional, and
biconditional in terms of the two given operators (without using truth tables).
Solution
The following logical equivalences express the logic operators disjunction, conditional and
biconditional in terms of only negation and conjunction.
p  q  (p  q)
p  q  (p q)
p  q  (p q)  (q p)
3. Define numerical DERIVE functions or, imp, and iff. Some values of these functions
should be:
or(0, 0) = 0
or(1, 1) = 1
imp(1, 0) = 0
iff(1, 0) = 0.
Hint: Use the definitions developed for the two preceding questions.
Verify that your functions yield correct results in all cases.
Solution
Author the following DERIVE expressions and simplify the last expression to show that
correct results are obtained for each of the three functions.
or(p, q) := not(and(not(p), not(q)))
imp(p, q) := not(and(p, not(q)))
iff(p, q) := and(not(and(p, not(q))), not(and(q, not(p))))
A := VECTOR(VECTOR([p, q, or(p, q), imp(p, q), iff(p, q)], q, 0, 1), p, 0, 1)
APPEND(A SUB 1, A SUB 2)
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