Propositional Logic Questions and Answers

Q: Let \(p\) and \(q\) be the following propositions:\(p\): Fail grade can be given.\(q\): Student scores more than 50% marks.Consider the statement: "Fail grade cannot be given when student scores more than 50% marks."Which one of the following is the CORRECT representation of the above statement in propositional logic?

The correct representation is \(q \to \neg p\).'Fail grade cannot be given when student scores more than 50% marks' means: if the student scores more than 50% (\(q\)), then a fail grade cannot be given (\(\neg p\)). The 'when' clause is the condition, so it goes on the left of the implication.\(q \to p\) says a fail grade can be given to such a student, the opposite. \(p \to q\) says that if a fail grade can be given then the student scored more than 50%, which reverses the direction and drops the negation. \(\neg p \to q\) is the converse of the correct statement, and an implication is not equivalent to its converse.

Q: The inference rule deals exclusively with formulas in conjunctive normal forms is called as :

The rule is the resolution rule.Resolution works on clauses, disjunctions of literals, so the formulas must first be put into conjunctive normal form. From two clauses containing a complementary pair, such as \(P \vee Q\) and \(\neg P \vee R\), it derives the resolvent \(Q \vee R\). Deriving the empty clause shows the clause set is unsatisfiable, which is how resolution refutation proves theorems.Contingency, unsatisfiability and contradiction are properties of formulas, not names of inference rules.

Q: A committee is to be selected from 5 candidates a, b, c, d and e. The selection must satisfy all the following conditions :Either a or b must be included, but not bothEither c or e or both must be includedIf d is included then b must be includedEither a or c are included or neither is included.If e is included, then c and d must be included. How should selection be made ?

The selection is \(a \wedge \bar{b} \wedge c \wedge \bar{d} \wedge \bar{e}\): choose a and c only.Check each condition with a and c chosen. Exactly one of a, b is in. At least one of c, e is in. d is out, so the rule 'if d then b' is not triggered. a and c are both in, which satisfies 'both or neither'. e is out, so the last rule is not triggered.Each other option leaves out both a and b, which breaks the first condition at once. They fail elsewhere too: with c in and a out, 'both or neither' fails; including e without d breaks the last rule; including d without b breaks the third.

Let \(p\) and \(q\) be the following propositions:

\(p\): Fail grade can be given.
\(q\): Student scores more than 50% marks.

Consider the statement: "Fail grade cannot be given when student scores more than 50% marks."

Which one of the following is the CORRECT representation of the above statement in propositional logic?

A.\(q \to \neg p\)
B.\(q \to p\)
C.\(p \to q\)
D.\(\neg p \to q\)

Exam Year(s)

GATE CS - 2024,

Answer: A

The correct representation is \(q \to \neg p\).

'Fail grade cannot be given when student scores more than 50% marks' means: if the student scores more than 50% (\(q\)), then a fail grade cannot be given (\(\neg p\)). The 'when' clause is the condition, so it goes on the left of the implication.

\(q \to p\) says a fail grade can be given to such a student, the opposite. \(p \to q\) says that if a fail grade can be given then the student scored more than 50%, which reverses the direction and drops the negation. \(\neg p \to q\) is the converse of the correct statement, and an implication is not equivalent to its converse.

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The inference rule deals exclusively with formulas in conjunctive normal forms is called as :

A.Resolution rule
B.Contingency rule
C.Unsatisfiability rule
D.Contradiction rule

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The rule is the resolution rule.

Resolution works on clauses, disjunctions of literals, so the formulas must first be put into conjunctive normal form. From two clauses containing a complementary pair, such as \(P \vee Q\) and \(\neg P \vee R\), it derives the resolvent \(Q \vee R\). Deriving the empty clause shows the clause set is unsatisfiable, which is how resolution refutation proves theorems.

Contingency, unsatisfiability and contradiction are properties of formulas, not names of inference rules.

Discuss in Forum

A committee is to be selected from 5 candidates a, b, c, d and e. The selection must satisfy all the following conditions :

Either a or b must be included, but not both
Either c or e or both must be included
If d is included then b must be included
Either a or c are included or neither is included.
If e is included, then c and d must be included. How should selection be made ?

A.\(a \wedge \bar{b} \wedge c \wedge \bar{d} \wedge \bar{e}\)
B.\(\bar{a} \wedge \bar{b} \wedge c \wedge \bar{d} \wedge \bar{e}\)
C.\(\bar{a} \wedge \bar{b} \wedge c \wedge \bar{d} \wedge e\)
D.\(\bar{a} \wedge \bar{b} \wedge c \wedge d \wedge \bar{e}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The selection is \(a \wedge \bar{b} \wedge c \wedge \bar{d} \wedge \bar{e}\): choose a and c only.

Check each condition with a and c chosen. Exactly one of a, b is in. At least one of c, e is in. d is out, so the rule 'if d then b' is not triggered. a and c are both in, which satisfies 'both or neither'. e is out, so the last rule is not triggered.

Each other option leaves out both a and b, which breaks the first condition at once. They fail elsewhere too: with c in and a out, 'both or neither' fails; including e without d breaks the last rule; including d without b breaks the third.

Discuss in Forum

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