Reasoning

Logic and Fallacies

1,716 Questions

Understand the fundamentals of propositional logic, logical inference rules, and paradoxes. This set includes identifying logical fallacies, including those found in classical Nyaya logic. Strong grasp of these concepts is crucial for scoring well in the reasoning sections of competitive tests.

Propositional logic modelsLogical inference rulesTypes of logical fallaciesParadoxes and contingent truthsNyaya logic concepts

Logic and Fallacies Questions

Multiple choice maths mathematical reasoning implications principle of mathematical induction proofs in mathematics

The proposition $(p\rightarrow \sim p)\wedge (\sim p\rightarrow p)$ is a

  1. tautology.

  2. contradiction.

  3. neither a tautology nor a contradiction.

  4. tautology and contradiction.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 $p$ $\sim p $  $p\rightarrow \sim p $ $\sim p \rightarrow p$  $(p\rightarrow \sim p) \wedge(\sim p\rightarrow p)$ 
T  F  F  T  F 
F  T  T  F  F 

A contradiction.

Multiple choice business maths mathematical reasoning implications principle of mathematical induction proofs in mathematics

$∼(p⇒q)⟺∼p\vee ∼q  \, is$

  1. a tautology

  2. a contradiction

  3. neither a tautology nor a contradiction

  4. cannot come to any conclusion
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression ~(p -> q) is equivalent to (p AND ~q). The expression (~p OR ~q) is the negation of (p AND q). These are not equivalent, so the statement is neither a tautology nor a contradiction.

Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Consider the following statements 
$p$:you want to success
$q$:you will find way,
then the negation of $\sim (p\vee q)$ is

  1. you want of success and you find a way

  2. you want of success and you do not find a way

  3. if you do not want to succeed then you will find a way

  4. if you want of success then you cannot find a way
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Which of the following statements is a tautology

  1. $\left( { \sim p \vee q} \right) - \left( {p \vee \sim q} \right)$
  2. $\left( { \sim p \vee \sim q} \right) \to p \vee q$
  3. $\left( {p \vee \sim q} \right) \wedge \left( {p \vee q} \right)$
  4. $\left( { \sim p \vee \sim q} \right) \vee \left( {p \vee q} \right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Which of the following is correct?

  1. $(~p \vee ~q) \equiv (p \wedge q)$
  2. $(p \rightarrow q) \equiv (~q \rightarrow ~p)$
  3. $~(p \rightarrow ~q) \equiv (p \wedge ~q)$

  4. none of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Clearly, the statements $p \vee q$ and $p\wedge q$ cannot be equivalent as they one operator means "OR" and the other operator means "AND".

$p$ $q$ $p\rightarrow q$ $q\rightarrow p$
T T T T
T F F T
F T T F
F F T T

Option B is also incorrect.

$p$ $q$ $p\rightarrow q$ $p\wedge q$
T T T T
T F F F
F T T F
F F T F


Hence, option C is also incorrect.

Option D is also incorrect as
$p \leftrightarrow q=(p\rightarrow q)\wedge (q\rightarrow p)$


Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Which of the following statement are NOT logically equivalent?

  1. $ \sim (p \vee \sim q)$ and $ (\sim p \wedge q )$
  2. $\sim (p \rightarrow q )$ and $(p \wedge \sim q )$
  3. $(p \rightarrow q) $ and $(\sim q \rightarrow \sim p) $
  4. $(p \rightarrow q )$ and $(\sim p \wedge q)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We make an option wise check for this.

Option A: $\sim \left( p\vee \sim q \right) \quad and\quad \left( \sim p\wedge q \right) $
By application of Demorgan's Law on $\sim \left( p\vee \sim q \right) $ we get, $\sim \left( p\wedge q \right) $ 
So this option is logically equivalent.

Option B: $\sim \left( p\longrightarrow q \right) \quad and\quad \left( p\wedge \sim q \right) $
Again by application Conditional Disjunction rule, we see that this option is also logically equivalent.

Option C: $\left( p\longrightarrow q \right) \quad and\quad \left( \sim q\longrightarrow \sim p \right) $
This is again true by Contrapositive tautology.

Option D:$\left( p\longrightarrow q \right) \quad and\quad \left( \sim p\wedge q \right) $
This is not logically equivalent. 

Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Which of the following is always true?

  1. $\sim(p\rightarrow q) \equiv \sim p \wedge q$
  2. $\sim(p\vee q) \equiv \sim p \vee \sim q$
  3. $\sim (p \implies q ) \equiv (p \land \sim q )$
  4. $\sim(p \wedge q) \equiv \sim p \wedge \sim q$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$p \implies q \equiv \sim p \lor q  $
$\therefore \sim (p \implies q ) \equiv \sim (\sim p \lor q )$
$\therefore \sim (p \implies q ) \equiv (p \land \sim q )$

Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Logically equivalent statement to $p \leftrightarrow  q$ is

  1. $(p \rightarrow q)\wedge (q \rightarrow p)$
  2. $(p \wedge q)\vee (q \rightarrow p)$
  3. $(p \wedge q)\rightarrow (q \vee p)$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
 $p$  $q$  $p\leftrightarrow q$
 T  T  T
 T  F  F
 F  T  F
 F  F  T
 $p$  $q$  $p\rightarrow q$  $q\rightarrow p$ $\left( p\longrightarrow q \right) \wedge \left( q\longrightarrow p \right) $ $p\wedge q$  $\left( p\wedge q \right) \vee \left( q\longrightarrow p \right) $ $q\vee p$  $\left( p\wedge q \right) \longrightarrow \left( q\vee p \right) $ 
 T  T  T  T  T  T  T  T  T
T   F  F  T  F  F  T  T  T
 F  T  T  F  F  F  F  T  T
 F F   T  T  T  F  T  F  T
Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Which of the following is NOT true for any two statements $p$ and $q$?

  1. $\sim[p\vee (\sim q)]=(\sim p)\wedge q$
  2. $\sim(p\vee q)=(\sim p)\vee (\sim q)$
  3. $q\wedge \sim q$ is a contradiction
  4. $\sim (p\wedge (\sim p))$ is a tautology
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$p$ and $q$ are two statements.
$A) LHS = \sim [pv (\sim q)]$
By De morgon's laws
$\sim(pr (\sim q))= \sim pnq$
$\therefore (A) $ is true .

$B) \sim(p v q) = (\sim p) \vee (\sim q)$
According to demorgon's laws, this is false.
$\because \sim (p \vee q) = (\sim p)\wedge (\sim q)$. 
$\therefore (B)$ is false.

$C) q \wedge \sim  q$ is a contradiction because $'q'$ and $\sim q$ are opposite statements i.e, cannot be there at the same time.

$D) \sim (p \wedge (\sim p))$
$p \wedge (\sim p)$ is a contradiction, which is evident from option $(C)$. $\therefore $ opposite of a contradiction is a tautology .
$\therefore [B]$ is wrong.