The statement (p∧q) ⇒ (p∧r) is equivalent to :
- q⇒ (p∧r)
- p⇒ (p∧r)
- (p∧r) ⇒ (p∧q)
- (p∧q) ⇒ r
The Correct Option is D
Approach Solution - 1
To determine the equivalent logical statement for \((p \land q) \Rightarrow (p \land r)\), we will use logical equivalences and analyze each option.
The implication \((p \land q) \Rightarrow (p \land r)\) can be rewritten using the definition of implication: \(A \Rightarrow B\) is equivalent to \(\neg A \lor B\).
This means:
\[ (p \land q) \Rightarrow (p \land r) \equiv \neg (p \land q) \lor (p \land r) \]
Simplifying \(\neg (p \land q) \lor (p \land r)\) involves the following steps:
- The expression \(\neg (p \land q)\) using De Morgan's Laws becomes \(\neg p \lor \neg q\).
- Thus, \(\neg (p \land q) \lor (p \land r)\) becomes \((\neg p \lor \neg q) \lor (p \land r)\).
- The distributive law allows us to further simplify this to:
- \((\neg p \lor \neg q \lor p) \land (\neg p \lor \neg q \lor r)\)
- We know that \((\neg p \lor p)\) is a tautology and simplifies to \(True\), so the expression reduces to:
- \(True \land (\neg q \lor r)\)
- Which simplifies finally to \(\neg q \lor r\)
- \(\neg q \lor r\) is equivalent to \(q \Rightarrow r\).
Therefore, the equivalent statement to \((p \land q) \Rightarrow (p \land r)\) is \((p \land q) \Rightarrow r\).
This matches the correct option: \((p \land q) \Rightarrow r\).
Approach Solution -2
| A | B | ||||||||
| P | q | r | p∧ q | p∧ r | A → B | q → B | p → B | B → A | A → r |
| T | T | T | T | T | T | T | T | T | T |
| T | F | T | F | T | T | T | T | F | T |
| F | T | T | F | F | T | F | T | T | T |
| F | F | T | F | F | T | T | T | T | T |
| T | T | F | T | F | F | F | F | T | F |
| T | F | F | F | F | T | T | F | T | T |
| F | T | F | F | F | T | F | T | T | T |
| F | F | F | F | F | T | T | T | T | T |
(p∧ q) ⇒ (p∧ r) is equivalent to (p∧ q) ⇒ r
So, the correct option is (D): (p∧q) ⇒ r
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Concepts Used:
Mathematical Reasoning
Mathematical reasoning or the principle of mathematical reasoning is a part of mathematics where we decide the truth values of the given statements. These reasoning statements are common in most competitive exams like JEE and the questions are extremely easy and fun to solve.
Types of Reasoning in Maths:
Mathematically, reasoning can be of two major types such as:
- Inductive Reasoning - In this, method of mathematical reasoning, the validity of the statement is examined or checked by a certain set of rules, and then it is generalized. The principle of mathematical induction utilizes the concept of inductive reasoning.
- Deductive Reasoning - The principle is the opposite of the principle of induction. Contrary to inductive reasoning, in deductive reasoning, we apply the rules of a general case to a provided statement and make it true for particular statements. The principle of mathematical induction utilizes the concept of deductive reasoning.