Tag: different forms of theoretical statements

Questions Related to different forms of theoretical statements

The compound proposition which is always false is:

  1. $\left(p \rightarrow q\right)\leftrightarrow \left( \sim q \rightarrow \sim p \right) $

  2. $\left[ \left( p\rightarrow q \right) \wedge \left( q\rightarrow r \right) \right]\rightarrow \left( p\rightarrow r \right) $

  3. $\left( \sim p\vee q \right) \leftrightarrow \left( p\wedge \sim q \right) $

  4. $p \rightarrow \sim p$


Correct Option: A

$p \wedge ( q \wedge r )$  is logically equivalent to

  1. $p \vee ( q \wedge r )$

  2. $( p \wedge q ) \wedge r$

  3. $( p \vee q ) \vee r$

  4. $p \rightarrow ( q \wedge r )$


Correct Option: B

If  $p$ and  $q$ are two simple proposition then  $p \rightarrow q$  is false when

  1. $p \text { is true and } q \text{ is true}$

  2. $p \text { is false and } q  \text{ is true}$

  3. $p \text { is true and } q \text{ is false}$

  4. both $p$ and $q$ are false


Correct Option: B

Let  $p :$  Mathematics is interesting and let  $q:$  Mathematics is difficult, then the symbol  $p\wedge q$  means

  1. Mathematics is interesting implies that Mathematics is difficult

  2. Mathematics is interesting implies and is implied by Mathematics is difficult

  3. Mathematics is interesting and Mathematics is difficult

  4. Mathematics is interesting or Mathematics is difficult


Correct Option: C
Explanation:

$'\Lambda '$ stands for logical and 

$\therefore$    $p\Lambda q$ means 
Mathematics is interesting and Mathematics is difficult.

The dual of the statement $\left[ p\wedge \left( \sim q \right)  \right] \wedge \left( \sim p \right)] $ is

  1. $p\vee \left( \sim q \right) \vee \sim p$

  2. $\left( p\vee \sim q \right) \vee \sim p$

  3. $p\wedge \sim \left( q\vee \sim p \right) $

  4. none of these


Correct Option: B

The contrapositive of the sentence $\sim p \rightarrow q$ is equivalent to

  1. $p \rightarrow \sim q$

  2. $q \rightarrow \sim p$

  3. $q \rightarrow p$

  4. $\sim p \rightarrow \sim q$

  5. $\sim q \rightarrow \sim p$


Correct Option: E
Explanation:

For a conditional statement p → q, Its converse statement (q → p) and inverse statement (∼p → ∼q) are equivalent to each other. p → q and its contrapositive statement (∼q → ∼p) are equivalent to each other.

Write the inverse and contrapositive of the statement
"If two triangles are congruent, then their areas are equal."
$(a)$Inverse of the statement :
If two triangles are not congruent, then their areas are equal.
$(b)$Contrapositive of the statement:
If the areas of the two triangles are equal, then the triangles are congruent.

  1. $(a)False$ and $(b)$ False

  2. $(a)True$ and $(b)$ False

  3. $(a)False$ and $(b)$ True

  4. $(a)True$ and $(b)$ True


Correct Option: A
Explanation:

"If two triangles are congruent, then their areas are equal."
$(a)$Inverse of the statement :
If two triangles are not congruent, then their areas are not equal.
$(b)$Contrapositive of the statement:
If the areas of the two triangles are not equal, then the triangles are not congruent.

What is the symbolic form and truth value of the following?
"If $4$ is an odd number, then $6$ is divisible by $3$." 
p: $4$ is an odd number.
q: $6$ is divisible by $3$.

  1. p$\rightarrow$q and $F$

  2. q$\rightarrow$p and $T$

  3. q$\rightarrow$p and $F$

  4. p$\rightarrow$q and $T$


Correct Option: D
Explanation:

$p: 4$ is an odd number.
$q: 6$ is divisible by $3$.
Symbolic form: $p$ $\rightarrow$ $q$
$p$ is false and $q$ is true.
So, $F\rightarrow T$ is $T$.

Identify the Law of Logic
$\sim(\sim p) \equiv p$

  1. DeMorgan's Law

  2. Conditional Law

  3. Involution Law

  4. Complement Law


Correct Option: C
Explanation:

Involution Law

This law states that if you negate a negation they effectively cancel each other out.
$\sim (\sim p)\equiv p$

Which of following is the negation of $(P \ \vee\sim Q).$

  1. $\sim P\vee Q$

  2. $\sim P\wedge Q$

  3. $\sim Q\wedge P$

  4. $\sim Q\vee P$


Correct Option: B
Explanation:
 P  Q  $\sim P$  $\sim Q$  $P\vee \sim Q$ $\sim \left( P\vee \sim Q \right) $  $\sim P\wedge Q$ 
 T  F  F  T  F  F
T  F  T  T  F  F
F  T  F  F  T  T
F F  T  T  F  T  F
Therefore, $\sim \left( P\vee \sim Q \right) $ is $\sim P\wedge Q$