Reasoning

Logic and Fallacies

1,803 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 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

$(p\rightarrow q)\leftrightarrow (q\vee \sim p)$ is - 

  1. Equivalent to $p\wedge q$

  2. Tautology

  3. Fallacy

  4. Neither tautology nor fallacy
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The implication p implies q is logically equivalent to not p or q. The given statement equates this with (q or not p), which is identical by the commutative law. Thus, both sides are always equal, making the bi-implication a tautology.

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

If p and q are two statements, then statement $p\Rightarrow q\wedge \sim q$.

  1. Tautology

  2. Contradiction

  3. Neither tautology nor contradiction

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In the statement p implies (q and not q), the consequent (q and not q) is always false (a contradiction). An implication with a false consequent and a variable antecedent has a truth value that depends on p, making it neither a tautology nor a contradiction.

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

The statement $\sim (p \leftrightarrow \sim q)$ is

  1. Equivalent to $\sim p \leftrightarrow q$

  2. A tautology

  3. A fallacy
  4. Equivalent to $p \leftrightarrow q$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The biconditional p <-> q means both have the same truth value, while p <-> not q means they have opposite truth values. Negating a biconditional that equates p to not q flips it back to equating p directly to q.

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

The proposition $\left( {p \wedge q} \right) \Rightarrow p$ is 

  1. neither tautology nor contradiction

  2. A tautology

  3. A contradiction

  4. Cannot be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The proposition (p and q) implies p means that whenever both p and q are true, p must be true, which is always correct by definition of conjunction and implication. Thus, it is a tautology.

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

The only statement among the following that is a tautology is-

  1. $A\wedge \left( A\vee B \right) $
  2. $A\vee \left( A\wedge B \right) $
  3. $[A\wedge (A\rightarrow B)]\rightarrow B$
  4. $B\rightarrow [A\wedge (A\vee B)]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

[A ^ (A -> B)] -> B is Modus Ponens, which is a tautology. If A is true and (A -> B) is true, then B must be true.