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Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

Which of the following is a contradiction?

  1. $p\vee q$
  2. $p\wedge q$
  3. $p\vee (\sim p)$
  4. $p\wedge (\sim p)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$p$  $q$  $p\vee q$  $p\wedge q$
$ T$ $ T$ $ T$ $ T$
$ T$ $ F$ $ T$ $ F$
$ F$ $T$ $T$ $F$
$ F$ $F$ $ F$ $ F$
So $p\vee q$ and $p\wedge q$ are not contradiction 

Now check other options:

| $p$ |  $\sim p$ |  $p\vee (\sim p)$ |  $p\wedge (\sim p)$ | | --- | --- | --- | --- | |  $T$ | $ F$ | $ T$ | $ F$ | | $ F$ | $T$ | $T$ | $F$      |
Hence $p\wedge (\sim p)$ is contradiction 

Note: If a compound statement is always False , then it is called contradiction 
Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

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

  1. a tautology

  2. a contradiction

  3. neither a tautology nor a contradiction

  4. a tautology and a contradiction
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$p$      $q$      $\sim p$      $p\rightarrow \sim p$     $\sim p \rightarrow q$     $(p\rightarrow \sim p)\wedge(\sim p\rightarrow q)$
T T    F      F       T         F
T F    F      F       T         F
F T    T      T       T         T
F F    T      T       F         F
Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

The statement $(p-q)\rightarrow [(\sim p \rightarrow q)\rightarrow q]$ is 

  1. a tautology
  2. equivalent to $\sim p \rightarrow q$
  3. equivalent to $p\rightarrow \sim q$

  4. a fallacy
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The statement (p -> q) -> ((~p -> q) -> q) is a tautology. It can be verified using a truth table or logical equivalences.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

Which of the following proposition is a contradiction?

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

Contradiction is a preposition which is always false(F).
Here $\sim \equiv negation$ and $\wedge \equiv AND$
Let $ x=(\sim p \wedge q) \wedge (\sim q)$
If $p=F$ and $q=F$ then $x=F$
If $p=F$ and $q=T$ then $x=F$
If $p=T$ and $q=F$ then $x=F$
If $p=T$ and $q=T$ then $x=F$
Hence option (c) is correct

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

$p,q,r$ are $3$ statement such that $(p\rightarrow q)\wedge (q\rightarrow r)\Rightarrow (p\rightarrow r)$ is 

  1. Tautology

  2. Contradiction
  3. $P\wedge q$
  4. $p\wedge (\sim q)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression (p -> q) ^ (q -> r) -> (p -> r) is the Law of Hypothetical Syllogism, which is a fundamental tautology in propositional logic.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

$p,q,r$ are $3$ statement such that $(p \rightarrow q)\wedge (q \rightarrow r)\Rightarrow (P \rightarrow r)$ is

  1. Tautology

  2. Contradiction
  3. $P \wedge q$
  4. $p \wedge (\sim q)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is identical to the previous question; the expression (p -> q) ^ (q -> r) -> (p -> r) is a tautology.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

Which of the following is a tautology?

  1. $p\wedge (\sim p)$
  2. $p\wedge c$
  3. $p\vee t$
  4. $p\wedge p$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A tautology is a compound statement that is always true regardless of the truth values of its components. For option C, the disjunction of any proposition p with a true statement t always results in true, satisfying the definition of a tautology.