Which of the following is logically equivalent? A. ¬p → (q → r) and q → (p ∨ r) B. (p → q) → r and p → (q → r) C. (p → q) → (r → s) and (p → r) → (q → s) Choose the correct answer from the options given below :
A only
Logical equivalence is a fundamental concept in propositional logic. Two propositional formulas are said to be logically equivalent if they have the same truth value under all possible assignments of truth values to their propositional variables. This means they represent the same logical meaning. We can check for logical equivalence using truth tables or by applying a sequence of known logical equivalences.
Let's examine each pair of expressions given in the question to determine if they are logically equivalent.
We can analyze this pair by converting the implications into disjunctions using the equivalence $A \rightarrow B \equiv \neg A \lor B$.
Comparing the simplified forms, $p \lor \neg q \lor r$ is the same as $p \lor \neg q \lor r$. Therefore, the two expressions in Pair A are logically equivalent.
Let's simplify these expressions using the equivalence $A \rightarrow B \equiv \neg A \lor B$.
Comparing the simplified forms, $(p \land \neg q) \lor r$ is not the same as $\neg p \lor \neg q \lor r$. Implication is not associative. For example, consider the case where $p$ is False, $q$ is False, and $r$ is False:
Since the truth values differ for this assignment (F vs T), the two expressions in Pair B are not logically equivalent.
Let's simplify these expressions using the equivalence $A \rightarrow B \equiv \neg A \lor B$.
Comparing the simplified forms, $(p \land \neg q) \lor \neg r \lor s$ is not the same as $(p \land \neg r) \lor \neg q \lor s$. For example, consider the case where $p$ is False, $q$ is True, $r$ is False, and $s$ is False:
Since the truth values differ for this assignment (T vs F), the two expressions in Pair C are not logically equivalent.
Based on the analysis:
Therefore, only Pair A is logically equivalent.
| Equivalence Name | Equivalence |
|---|---|
| Double Negation | $\neg (\neg p) \equiv p$ |
| Implication | $p \rightarrow q \equiv \neg p \lor q$ |
| De Morgan's Laws | $\neg (p \land q) \equiv \neg p \lor \neg q$ $\neg (p \lor q) \equiv \neg p \land \neg q$ |
| Commutative Laws | $p \land q \equiv q \land p$ $p \lor q \equiv q \lor p$ |
| Associative Laws | $(p \land q) \land r \equiv p \land (q \land r)$ $(p \lor q) \lor r \equiv p \lor (q \lor r)$ |
| Distributive Laws | $p \land (q \lor r) \equiv (p \land q) \lor (p \land r)$ $p \lor (q \land r) \equiv (p \lor q) \land (p \lor r)$ |
Propositional logic is a branch of logic that studies ways of joining and/or modifying entire propositions, statements, or sentences to form more complicated propositions, statements, or sentences, as well as the logical relationships and properties that are derived from these methods of combining or altering statements. Key components include:
Understanding logical equivalence is crucial for simplifying logical expressions, proving theorems, and designing logic circuits.
A two-input logic gate is giving high output only when both the inputs are high. For all other input conditions, the output is low. Select the correct logic gate.
The output is high only if one of the input is high. The above statement represents _____
The number of gate inputs, required to realize expression ABC + AB̅CD + EF̅ + AD is
Consider the expression Y = P ⨁ Q ⨁ R where P, Q, R are the input variables and Y is the output variable. Y will be logic 0 if
(A) an odd number of input variables are 1
(B) an even number of input variables are 1
(C) an odd number of inputs variables are 0
(D) an even number of input variable are 0
(E) an odd number of input variable between 0 and 1
Choose the correct answer from the options given below:
Match List I with List II
List – I | List – II | |||||||||||||||||
Logic gate | Truth table | |||||||||||||||||
A. | Negative OR | I. |
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B. | Exclusive NOR | II. |
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C. | Exclusive OR | III. |
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D. | Negative AND | IV. |
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Choose the correct answer from the options given below: