Kenneth Rosen Edition 7 Exercise 1.3 Question 35 (Page No. 35)

Question

Find the dual of each of these compound propositions.

1. $p \wedge \sim q \wedge \sim r$
2. $(p \wedge q \wedge r) \vee s$
3. $(p \vee F) \wedge (q \vee T)$

Answer 1

Dual is same as de morgans low . Except we can not complement  of literls in dual .

Comments

Literal$:-$  every occurrence of a variable in its true form $(or)$ complemented form.

Example$: A.\overline{B}+\overline{A}.B$

                 Number of literals$=4$

Answer 2

Dual = "The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T by F, and each F by T. The dual of s is denoted by s∗."

we don't negate any of the literal.

Answer:

a)  p$\vee$∼q$\vee$∼r

b)  (p$\vee$q$\vee$r)$\wedge$s

c)  (p$\wedge$T)$\vee$(q$\wedge$F)

Answer 3

The dual of a compound proposition is formed by replacing each occurrence of the logical operator "and" with "or", each occurrence of "or" with "and", and negating all propositions.

The compound proposition is : p ∧ ¬q ∧ ¬r

The dual of this compound proposition is ; p ∨ q ∨ r

(p ∧ q ∧ r) ∨ s

The dual of this compound proposition is: (p ∨ ¬q ∨ ¬r) ∧ ¬s

(p ∨ F) ∧ (q ∨ T)

The dual of this compound proposition is: (¬p ∧ ¬F) ∨ (¬q ∧ ¬T)

Note that ; The logical operator "and" is replaced with "or" and the negation of q and r is removed, since dual of a negation is the removal of the negation.