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:
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