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

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 )

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 )

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

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

MEDIUM

Earn 100

The statement $(\sim p \land q) \lor \sim q$ is

50% studentsanswered this correctly

Important Questions on Mathematical Reasoning

MEDIUM

Mathematics>Algebra>Mathematical Reasoning>Algebra of Statements

The statement pattern

(∽ p \land q)

is logically equivalent to

MEDIUM

Mathematics>Algebra>Mathematical Reasoning>Algebra of Statements

The expression

~ (~ p \to q)

is logically equivalent to

EASY

Mathematics>Algebra>Mathematical Reasoning>Algebra of Statements

The compound statement

(P \lor Q) \land (~ P) \Rightarrow Q

equivalent to:

EASY

Mathematics>Algebra>Mathematical Reasoning>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?

MEDIUM

Mathematics>Algebra>Mathematical Reasoning>Algebra of Statements

The negation of the Boolean expression

x \leftrightarrow ~ y

is equivalent to:

MEDIUM

Mathematics>Algebra>Mathematical Reasoning>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: