If the truth value of the statement
(P∧(∼R))→((∼R)∧Q)
is F, then the truth value of which of the following is F?
(P∧(∼R))→((∼R)∧Q)
is F, then the truth value of which of the following is F?
- P∨Q→∼R
- R∨Q→∼P
- ∼(P∨Q)→∼R
- ∼(R∨Q)→∼P
The Correct Option is D
Solution and Explanation
The correct answer is(D): ∼(R∨Q)→∼P.
\(\underbrace{P∧(∼R)X→}\) \(\underbrace{((∼R)∧Q)Y }\) = False
x y
X→Y = False
| X | Y | X→Y |
| F | F | T |
| T | T | T |
| F | T | T |
| T | F | F |
P∧∼R = T and (∼R)∧Q = F
⇒ P = T
∼R = T ⇒ R = F
⇒ P = T, Q = F and R = F
T∧Q = F
⇒ Q = F
Now ∼(R∨Q)→∼P
∼(F∨F)→∼F
F→F = False
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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.