Write the dual of p ∧ ~ p ≡ F .

To find the dual of p ∧ ~ p ≡ F In dual ∧ → ∨ , ∨ → ∧ , F → T , T → F Now let us make a truth table: p ~ p p ∧ ~ p T F F F T F Now, let us make truth table of p ∨ ~ p : p ~ p p ∨ ~ p T F T F T T Which gives p ∨ ~ p ≡ T Therefore, the dual of p ∧ ~ p ≡ F is p ∨ ~ p ≡ T

Write the dual of the following statement; ~ p ∧ ( q ∨ c ) .

The statement is ~ p ∧ ( q ∨ c ) Here, ∨ is the sign of disjunction, ∧ is the sign of conjunction and c is contradiction. To find the dual of the given statement, replace the conjunction sign ∧ with disjunction ∨ and disjunction sign ∨ with conjunction sign ∧ . And if there is any negation in the statement, do not change the negation of the statement. The duality of contradiction is tautology. So on replacing the connectives with each other, we get: ~ p ∨ ( q ∧ c )

The negation of the Boolean expression x ↔ ~ y is equivalent to:

( x ∧ y ) ∨ ( ~ x ∧ ~ y )

( ~ x ∧ y ) ∨ ( ~ x ∧ ~ y )

( x ∧ ~ y ) ∨ ( ~ x ∧ y )

( x ∧ y ) ∧ ( ~ x ∨ ~ y )

Let p , q , r be three logical statements. Consider the compound statements S 1 : ~ p ∨ q ∨ ~ p ∨ r and S 2 : p → q ∨ r Then, which of the following is NOT true?

If S 2 is False, then S 1 is True

If S 2 is True, then S 1 is True

If S 2 is False, then S 1 is False

If S 1 is False, then S 2 is False

Let F 1 A , B , C = A ∧ ~ B ∨ ~ C ∧ A ∨ B ∨ ~ A and F 2 A , B = A ∨ B ∨ B → ~ A be two logical expressions. Then :

F 1 is not a tautology but F 2 is a tautology

F 1 is a tautology but F 2 is not a tautology

Both F 1 and F 2 are not tautologies

F 1 and F 2 both are tautologies

EASY

Earn 100

Using the rules of negation write the negation of the statement $~ p \dot{\lor} (q \to\sim r)$ .

Important Questions on Mathematical Logic

EASY

Mathematics>Algebra>Mathematical Logic>Algebra of Statements

Write the dual of

p \land ~ p \equiv F

MEDIUM

Mathematics>Algebra>Mathematical Logic>Algebra of Statements

The Boolean expression

((p \land q) \lor (p \lor ~ q)) \land (~ p \land ~ q)

is equivalent to

EASY

Mathematics>Algebra>Mathematical Logic>Algebra of Statements

Write the dual of the following statement;

$~ p \land (q \lor c)$ .

MEDIUM

Mathematics>Algebra>Mathematical Logic>Algebra of Statements

The statement pattern

(∽ p \land q)

is logically equivalent to

MEDIUM

Mathematics>Algebra>Mathematical Logic>Algebra of Statements

The negation of the Boolean expression

x \leftrightarrow ~ y

is equivalent to:

EASY

Mathematics>Algebra>Mathematical Logic>Algebra of Statements

Write the dual of the following statement;

"Shweta is a doctor or Seema is a teacher"

MEDIUM

Mathematics>Algebra>Mathematical Logic>Algebra of Statements

If the truth value of the Boolean expression $((p \lor q) \land (q \to r) \land (~ r)) \to (p \land q)$ is false, then the truth values of the statements $p, q, r$ respectively can be:

EASY

Mathematics>Algebra>Mathematical Logic>Algebra of Statements

The negation of the Boolean expression

p \lor (~ p \land q)

is equivalent to :

EASY

Mathematics>Algebra>Mathematical Logic>Algebra of Statements

The dual of the statement pattern

~ p \land (q \lor t)

is (where

t

is a tautology and

c

is a contradiction)

MEDIUM

Mathematics>Algebra>Mathematical Logic>Algebra of Statements

The Boolean expression

(p \Rightarrow q) \land (q \Rightarrow ~ p)

is equivalent to :

EASY

Mathematics>Algebra>Mathematical Logic>Algebra of Statements

Let $p, q, r$ be three logical statements. Consider the compound statements
$S_{1} : ((~ p) \lor q) \lor ((~ p) \lor r)$ and $S_{2} : p \to (q \lor r)$
Then, which of the following is NOT true?