MATHEMATICAL REASONING (Logical Statements, Tautology and Contradiction)
Mathematical logic
Statement (Proposition): A sentence which is either true or false but not both is called a statement. They are denoted by $\mathrm{p}, \mathrm{q}, \mathrm{r}, \mathrm{s},………..$, A true statement is called a valid statement. If a statement is false we call it invalid statement.
Truth table
It is a tabular device to obtain the truth value of a compound statement or to check the validity of a simple or compound statement
Number of horizontal lines in a truth table depends on the number of substatements present in it.
If the problem involves $\mathrm{n}$ simple statements then number of rows is $2^{\mathrm{n}}$.
i.e.
| No. of statements | No. of row |
|---|---|
| 1 | $2^{1}=2$ |
| 2 | $2^{2}=4$ |
| 3 | $2^{3}=8$ |
| 4 | $2^{4}=16$ |
Logical connectives
(i). Conjunction $(\wedge)$
A compound statement joining two statements by “and” is called a conjunction and is denoted by $\wedge$.
i.e. the conjunction of two statements $p$ and $q$ is denoted by $p \wedge q . p \wedge q$ is true only if both the components $p$ and $q$ are true otherwise it is false.
(ii). Disjunction ( $\vee$ )
A compound statement joining two statements by “or” is called a disjunction and is denoted by $\vee$.
i.e. the disjunction of two statements $p ~ \& ~ q$ is true or both $p$ and $q$ is true otherwise $p \vee q$ is false.
We can express both $\mathrm{p} \vee \mathrm{q}$ and $\mathrm{p} \vee \mathrm{q}$ is tabular form as under. $\mathrm{T}$ stands for true and $\mathrm{F}$ stands for false.
| $\mathrm{p}$ | $\mathrm{q}$ | $\mathrm{p} \wedge \mathrm{q}$ | $\mathrm{p} \vee \mathrm{q}$ |
|---|---|---|---|
| $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{T}$ |
| $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{T}$ |
| $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{T}$ |
| $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{F}$ |
Quantifiers : phrases like “there exists” $(\exists)$ and “for all” $(\forall)$ are called quantifiers.
(iii). Implications $(\Rightarrow)$
$\mathrm{p} \Rightarrow \mathrm{q}$. One way implication
Here $\mathrm{p}$ is called antecedent or hypothesis or premise and $q$ is called consequence or conclusion.
$p \Rightarrow q$ is the same for each of the following
(i). If $\mathrm{p}$ then $\mathrm{q}$
(ii). $\mathrm{P}$ is sufficient for $\mathrm{q}$
(iii). $q$ is necessary condition for $p$
(iv). $p$ only if $q$
(v). $q$ if $p$
(vi). $q$ follows from $p$
(vii). $q$ is consequence of $p$
Since a true statement cannot imply a false statement, $p \Rightarrow q$ is always true except when $p$ is true and $q$ is false. It may also be useful to note that $p \Rightarrow q$ is equivalent to $\sim p \vee q$
Note : The contrapositive of a conditional statement is formed by negating both the hypothesis and the conclusion and then interchanging the resulting negations.
In otherwards, the contrapositivse negates and switches the parts of the sentence. It does both the jobs of the INVERSE and the CONVERSE.
IMPORTANT : Contrapositive has the same truth value as the original conditional statement.
Note :
| Statement | Converse | Inverse | Contrapositive | Negation |
|---|---|---|---|---|
| $\mathrm{p} \Rightarrow \mathrm{q}$ | $\mathrm{q} \Rightarrow \mathrm{p}$ | $\sim \mathrm{p} \Rightarrow \sim \mathrm{q}$ | $\sim \mathrm{q} \Rightarrow \sim \mathrm{p}$ | $\sim(\mathrm{p} \Rightarrow \mathrm{q})$ |
The inverse and the converse of a conditional statement are logically equivalent to each other, just as the conditional and its contrapositive are logically equivalent to each other.
(iv). Two way implication $(\Leftrightarrow)$
$p \Leftrightarrow q$, $“p$ implies and implied by $q”$ or $“p$ if and only if $q”$
$p \Leftrightarrow q$ is true if both $p \& q$ are true or both false and false when one of the statements is true and the other is false.
| $\mathrm{p}$ | $\mathrm{q}$ | $\mathrm{p} \Rightarrow \mathrm{q}$ | $\mathrm{q} \Rightarrow \mathrm{p}$ | $\mathrm{p} \Leftrightarrow \mathrm{q}$ |
|---|---|---|---|---|
| $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{T}$ |
| $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{F}$ |
| $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{F}$ |
| $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{T}$ |
(v). Negation
If $\mathrm{p}$ is a statement then negation $\mathrm{p}$ is written as $\sim \mathrm{p}$.
If $p$ is true then $\sim p$ is false. If $p$ is false then $\sim p$ is true
Note :
| $\mathrm{p}$ | $\sim \mathrm{p}$ |
|---|---|
| $\mathrm{T}$ | $\mathrm{F}$ |
| $\mathrm{F}$ | $\mathrm{T}$ |
-
$\sim(\sim \mathrm{p})=\mathrm{p}$
-
$\sim(p \wedge q)=\sim p \vee \sim q$
-
$\sim(p \vee q)=\sim p \wedge \sim q$
-
$\sim(p \Rightarrow q)=(p \wedge \sim q)$
-
$\sim(p \Leftrightarrow q)=(p \wedge \sim q) \vee(\sim p \wedge q)$
(vi). $\operatorname{NOR}(\downarrow)$
Let $\mathrm{p} \& \mathrm{q}$ be two statements. Then " $\mathrm{p} \downarrow \mathrm{q}$ " is called Joint Denial or “NOR” statement
(combination of NOT and OR) and read as “Neither $p$ nor $q”. “p \downarrow q$ " can also be written as $\neg(p \vee q)$ or $\sim(p \vee q)$.
Joint Denial is true only when $\mathrm{p}$ and $q$ both are false.
| $\mathrm{p}$ | $\mathrm{q}$ | $\mathrm{p} \downarrow \mathrm{q}=\sim(\mathrm{p} \wedge \mathrm{q})$ |
|---|---|---|
| $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{F}$ |
| $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{F}$ |
| $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{F}$ |
| $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{T}$ |
(vii). NAND ( $\uparrow)$
Let $\mathrm{p} \& \mathrm{q}$ be two statements. Then " $\mathrm{p} \uparrow q$ " is called NAND statement (combination of NOT and AND) and is written as " $\mathrm{p} \uparrow q$ "
" $\mathrm{p} \uparrow \mathrm{q}$ " is also written as $\rceil(\mathrm{p} \wedge \mathrm{q})$ or $\sim(\mathrm{p} \wedge \mathrm{q})$ This statement is false only if both $\mathrm{p} \& \mathrm{q}$ are true.
| $\mathrm{p}$ | $\mathrm{q}$ | $\mathrm{p} \uparrow \mathrm{q}=\sim(\mathrm{p} \vee \mathrm{q})$ |
|---|---|---|
| $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{F}$ |
| $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{T}$ |
| $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{T}$ |
| $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{T}$ |
Here is a table that shows a commonly used Precedence of logical operators
| Operator | Precedence |
|---|---|
| $\sim$ | 1 |
| $\wedge$ | 2 |
| $\vee$ | 3 |
| $\Rightarrow$ | 4 |
| $\Leftrightarrow$ | 5 |
Use of brackets
(i). If negation (i.e $\rceil$ or $\sim$ ) is repeated in the same statement then there is no need of bracket.
(ii). If in a statement, the connectives of same type are present, then brackets are applied from left.
(iii). If different connectives are used in a statement, then we remove the bracket of lower order connective. But we cannot remove the bracket of higher order connective.
For example:
(i). $\mathrm{p} \Rightarrow(\mathrm{q} \wedge \mathrm{r})=\mathrm{p} \Rightarrow \mathrm{q} \wedge \mathrm{r} \quad$ (order of $\wedge$ is less than order of $\Rightarrow$ )
(ii). $\mathrm{p} \vee(\mathrm{q} \Rightarrow \mathrm{r}) \neq \mathrm{p} \vee \mathrm{q} \Rightarrow \mathrm{r}$
Table of Symbols
| (i). If $p$ then $q$ | $p \Rightarrow q$ |
| (ii). $\mathrm{p}$ if $q$ | $q \Rightarrow p$ |
| (iii). $p$ only if $q$ | $q \rightarrow p$ |
| (iv). $\mathrm{p}$ unless $\mathrm{q}$ | $\sim q \Rightarrow p$ |
| (v). $\mathrm{p}$ is a sufficient condition for $q$ | $p \Rightarrow q$ |
| (vi). $p$ is a necessary condition for $q$ | $\mathrm{q} \Rightarrow \mathrm{p}$ |
| (vii). A sufficient condition for $p$ is $q$ | $q \Rightarrow p$ |
| (viii). A necessary condition for $\mathrm{p}$ is $\mathrm{q}$ | $p \Rightarrow q$ |
| (ix). In order that $p$ is sufficient that $q$ | $\mathrm{q} \Rightarrow \mathrm{p}$ |
| (x). In order that $p$ is necessary that $q$ | $P \Leftrightarrow q$ |
| (xi). $p$ if and only if $q$ | $p \rightarrow q$ |
| (xii). $\mathrm{p}$ is a necessary and sufficient condition for $\mathrm{q}$ | $\mathrm{p} \Leftrightarrow \mathrm{q}$ |
Tautology and contradiction (Fallacy)
A statement whose truth value is always T (i.e. True) is called a tautology and the statement whose truth value is always F(i.e.False) is called a contradiction. Negation of tautology is a contradiction and vice versa.
Logical equivalence
Two compound statements are said be logically equivalent if both have same truth values for all possible assignments given to the variables.
Duality
Two compounds are said to be dual of each other if either can be obtained from the other by interchanging $\wedge$ and $\vee$ provided both remain valid.
For e.g. the dual of $(p \vee q) \wedge r$ is $(p \wedge q) \vee r$.
Algebra of statements
| Commutative laws | (i). $(p \vee q) \Leftrightarrow(q \vee p)$ (ii). $(\mathrm{p} \wedge \mathrm{q}) \Leftrightarrow(\mathrm{q} \wedge \mathrm{p})$ |
| Associative laws | (i). $\mathrm{p} \vee(\mathrm{q} \vee \mathrm{r}) \Leftrightarrow(\mathrm{p} \vee \mathrm{q}) \vee \mathrm{r}$ (ii). $\mathrm{p} \wedge(\mathrm{q} \wedge \mathrm{r}) \Leftrightarrow(\mathrm{p} \wedge \mathrm{q}) \wedge \mathrm{r}$ |
| Distributive laws | (i). $\mathrm{p} \wedge(\mathrm{q} \vee \mathrm{r}) \Leftrightarrow(\mathrm{p} \wedge \mathrm{q}) \vee(\mathrm{p} \wedge \mathrm{r})$ (ii). $\mathrm{p} \vee(\mathrm{q} \wedge \mathrm{r}) \Leftrightarrow(\mathrm{p} \vee \mathrm{q}) \wedge(\mathrm{p} \vee \mathrm{r})$ |
| Idempotent laws | (i). $(p \vee p) \Leftrightarrow p$ (ii). $(\mathrm{p} \wedge \mathrm{p}) \Leftrightarrow \mathrm{p}$ |
| Abosorption laws | (i). $\mathrm{p} \vee(\mathrm{p} \wedge \mathrm{q}) \Leftrightarrow \mathrm{p}$ (ii). $\mathrm{p} \wedge(\mathrm{p} \vee \mathrm{q}) \Leftrightarrow \mathrm{p}$ |
| De Morgan’s laws | (i). $\sim(\mathrm{p} \vee \mathrm{q}) \Leftrightarrow(\sim \mathrm{p} \wedge \sim \mathrm{q})$ (ii). $\sim(\mathrm{p} \wedge \mathrm{q}) \Leftrightarrow(\sim \mathrm{p} \vee \sim \mathrm{q})$ |
| Detachment law | $((p \Rightarrow q) \wedge p) \Rightarrow q$ |
| Chain law | $((p \Rightarrow q) \wedge(q \Rightarrow p)) \Rightarrow(p \Rightarrow r)$ |
| Identity laws (t is tautology and $\mathrm{f}$ is contradiction) |
(i). $\mathrm{p} \wedge \mathrm{t}=\mathrm{t} \wedge \mathrm{p}=\mathrm{p}$ (ii). $p \vee f=f \vee p=p$ |
| Compliment laws | (i). $\mathrm{p} _{\vee} \sim \mathrm{p}=\mathrm{t}$ (ii). $\mathrm{p} _{\wedge} \sim \mathrm{p}=\mathrm{t}$ |
Solved examples
1. If $p \Rightarrow(q \vee r)$ is false, then the truth values of $p, q, r$ are respectively
(a). T, F, F
(b). F, T, T
(c). $\mathrm{T}, \mathrm{T}, \mathrm{F}$
(d). F, F, F
Show Answer
Solution:
$p \Rightarrow q$ is false only when $p$ is true and $q$ is false
$\mathrm{p} \Rightarrow(\mathrm{q} \vee \mathrm{r})$ is false when $\mathrm{p}$ is true and $\mathrm{q} \vee \mathrm{r}$ is false and $\mathrm{q} \vee \mathrm{r}$ is false when both $\mathrm{q}$ and $\mathrm{r}$ are false Hence, truth values of $p, q, r$ are respectively T, F, F.
Answer: (a)
2. Test the validity of the argument $\left(\mathrm{S} _{1}, \mathrm{~S} _{2} ; \mathrm{S}\right)$, where
$\mathrm{S} _{1} ; \mathrm{p} \vee \mathrm{q}, \mathrm{S} _{2}: \sim \mathrm{p}$ and $\mathrm{S}: \mathrm{q}$
Show Answer
Solution:
In order to test the validity of the argument $\left(\mathrm{S} _{1}, \mathrm{~S} _{2} ; \mathrm{S}\right)$, we first construct the truth table for the conditional statement.
$\mathrm{S} _{1} \wedge \mathrm{S} _{2} \rightarrow \mathrm{S}$ i.e $[(\mathrm{p} \vee \mathrm{q}) \wedge \sim \mathrm{p}] \rightarrow \mathrm{q}$
The truth table is as given below:
| $\mathrm{p}$ | $\mathrm{q}$ | $\mathrm{S} _{1}=\mathrm{p} \vee \mathrm{q}$ | $\mathrm{S} _{2}=\sim \mathrm{p}$ | $\mathrm{S} _{1} \wedge \mathrm{S} _{2}$ | $\mathrm{~S}=\mathrm{q}$ | $\mathrm{S} _{1} \wedge \mathrm{S} _{2} \rightarrow \mathrm{S}$ i.e. $\mathrm{S} _{1} \wedge \mathrm{S} _{2} \rightarrow \mathrm{q}$ |
|---|---|---|---|---|---|---|
| $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{T}$ |
| $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{T}$ |
| $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{T}$ |
| $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{T}$ |
We observe that the last comumn of the truth table for $\mathrm{S} _{1} \wedge \mathrm{S} _{2} \rightarrow \mathrm{S}$ contains $\mathrm{T}$ only. Thus, $\mathrm{S} _{1} \wedge \mathrm{S} _{2} \rightarrow$ $S$ is a tautology.
Hence, the given argument is valid.
3. If $(p \wedge \sim q) \wedge(\sim q \wedge q)$ is
(a). a contradiction
(b). a tautology
(c). neither a tautology nor a contradiction
(d). both a tautology and a contradiction
Show Answer
Solution:
$ \begin{aligned} (p \wedge \sim q) & \wedge(\sim p \wedge q) \equiv \quad(p \wedge \sim p) \wedge(\sim q \wedge q) \\ & \equiv(f \wedge f) \\ & \equiv f \end{aligned} $
$(\mathrm{f} \rightarrow$ false$)$
(By using associative laws and commutative laws)
$\therefore(p \wedge \sim q) \wedge(\sim p \wedge q)$ is a contradiction.
Answer: (a)
4. Which of the following is logically equivalent to $\sim(\sim p \Rightarrow q)$ ?
(a). $\sim \mathrm{p} \wedge \mathrm{q}$
(b). $\mathrm{p} \wedge \mathrm{q}$
(c). $\sim p _{\wedge} \sim \mathrm{q}$
(d). $\mathrm{p} _{\wedge} \sim \mathrm{q}$
Show Answer
Solution:
Since $\sim(p \Rightarrow q)=p \wedge \sim q$
$\sim(\sim p \Rightarrow q)=p _{\wedge} \sim q$
Answer: (c)
5. If $\mathrm{p} \equiv \mathrm{He}$ is a carpenter and $\mathrm{q}=\mathrm{He}$ is making a table.
Then write down the following statement into symbols:
(i). He is a carpenter and making a table.
(ii). He is a carpenter but is not making a table.
(iii). It is false that he is a carpenter or making a table.
(iv). Neither he is a carpenter nor he is making a table.
(v). He is not a carpenter and he is making a table.
(vi). It is false that he is not a carpenter or is not making a table.
(vii). He is a carpenter or making a table.
Show Answer
Solution:
The solution of above compound statements in terms of $p$ and $q$ are given below :
(i). $\mathrm{p} \wedge \mathrm{q}$
(ii). $\mathrm{p} \wedge \rceil\mathrm{q}$
(iii). $\rceil(\mathrm{p} \vee \mathrm{q})$
(iv). $\rceil \mathrm{p} \wedge\rceil \mathrm{q}$
(v). $\rceil \mathrm{p} \wedge \mathrm{q}$
(vi). $(\rceil \mathrm{p} \vee \rceil \mathrm{q})$
(vii). $\mathrm{p} \vee \mathrm{q}$
6. Write in words the converse, inverse, contrapositive and negation of the implication “If 2 is less then 3 , than $1 / 3$ is less than $1 / 2$.
Show Answer
Solution:
Let $\mathrm{p} \equiv 2$ is less than $3, \mathrm{q} \equiv 1 / 3$ is less than $1 / 2$.
Then implication is $p \Rightarrow q$ :
(i). Converse of $p \Rightarrow q$ is $q \Rightarrow p$. In words $q \Rightarrow p$ means “If $1 / 3$ is less than $1 / 2$, then 2 is less than 3 “.
(ii). Inverse of $p \Rightarrow q$ is $\rceil p \Rightarrow\rceil q$. Thus in words, $\rceil p \Rightarrow\rceil q$ means “If 2 is not less than 3 , then $1 / 3$ is not less than $1 / 2$ “.
(iii). Contrapositive of $p \Rightarrow q$ is $\rceil q \Rightarrow\rceil p$. Thus in words $\rceil q \Rightarrow\rceil p$ means “If $1 / 3$ is not less than $1 / 2$, then 2 is not less than 3 “.
(iv). Negation of $p q$ is $\rceil(p \Rightarrow q)$. Thus in words $\rceil$ ( $p \Rightarrow$ means “It is false than $p$ implies q”
7. The statement $\mathrm{p} \rightarrow(\mathrm{q} \rightarrow \mathrm{p})$ is equivalent to
(a). $\mathrm{p} \rightarrow(\mathrm{p} \rightarrow \mathrm{q})$
(b). $\mathrm{p} \rightarrow(\mathrm{p} \vee \mathrm{q})$
(c). $\mathrm{p} \rightarrow(\mathrm{p} \wedge \mathrm{q})$
(d). $\mathrm{p} \rightarrow(\mathrm{p} \wedge \mathrm{q})$
Show Answer
Solution:
$\mathrm{p} \rightarrow(\mathrm{q} \rightarrow \mathrm{p})=\sim \mathrm{p} \vee(\mathrm{q} \rightarrow \mathrm{p})$
$=\sim p \vee(\sim q \vee p)$ since $p \vee \sim p$ is always true
$=\sim \mathrm{p} \vee \mathrm{p} \vee \mathrm{q}=\mathrm{p} \rightarrow(\mathrm{p} \vee \mathrm{q})$
Answer: (b)
8. Statement - 1:
$\sim(\mathrm{p} \leftrightarrow \sim \mathrm{q})$ is equivalent to $\mathrm{p} \leftrightarrow \mathrm{q}$.
Statement - 2:
$(\sim \mathrm{p} \leftrightarrow \sim \mathrm{q})$ is a tautology.
(a). Statement - 1 is True, Statement - 2 is True ; Statement - 2 is a correct explanation for Statement - 1
(b). Statement - 1 is True, Statement - 2 is True; Statement - 2 is NOT a correct explanation for Statement - 1
(c). Statement - 1 is True, Statement - 2 is False
(d). Statement - 1 is False, Statement - 2 is True
Show Answer
Solution: :
| $\mathrm{p}$ | $\mathrm{q}$ | $\mathrm{p} \leftrightarrow \mathrm{q}$ | $\sim \mathrm{p}$ | $\mathrm{p} \leftrightarrow \sim \mathrm{q}$ | $\sim(\mathrm{p} \leftrightarrow \sim \mathrm{q})$ | $\sim \mathrm{q}$ | $\sim \mathrm{p} \leftrightarrow \sim \mathrm{q}$ |
|---|---|---|---|---|---|---|---|
| $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{T}$ |
| $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{F}$ |
| $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{F}$ |
| $\mathrm{F}$ | $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{F}$ | $\mathrm{T}$ | $\mathrm{T}$ | $\mathrm{T}$ |
Answer: (c)
Exercise
1. Are the following statements equivalent:
“If the traders do not reduce the price then the government will take action against them”. “It is not true that the traders do not reduce the prices and government does not take action against them”
Show Answer
Answer: Yes2. Which of the following is false?
(a). $(p \Rightarrow q) \Leftrightarrow(\sim q \Rightarrow \sim p)$ is a contradiction
(b). $\left(p _{\vee} \sim p\right)$ is a tautology
(c). $\sim(\sim p) \Leftrightarrow p$ is a tautology
(d). $(\mathrm{p} \wedge \sim \mathrm{p})$ is a contradiction
Show Answer
Answer: (a)3. If each of the statement $\mathrm{p} \Rightarrow \sim \mathrm{q} ; \mathrm{q} \Rightarrow \mathrm{r} ; \sim \mathrm{r}$ is true, then
(a). $\mathrm{p}$ is false
(b). $\mathrm{p}$ is true
(c). $q$ is true
(d). None of these
Show Answer
Answer: (a)4. Which of the following is true?
(a). $\sim(p \vee(\sim q)) \equiv(\sim p) \wedge q$
(b). $(p \wedge q) \wedge(\sim q)$ is a tautology
(c). $\sim(p \wedge(\sim p))$ is a contradiction
(d). None of these
Show Answer
Answer: (a)5. Which of the following is the contrapositive of ‘If two triangles are identical, then these are similar’?
(a). If two triangles are not similar, then these are not identical
(b). If two triangles are not identical, then these are not similar
(c). If two triangles are not identical, then these are similar
(d). If two triangles are not similar, then these are identical
Show Answer
Answer: (a)6. The contrapositive of the converse of $\mathrm{p} \Rightarrow \sim \mathrm{q}$ is
(a). $\sim q \Rightarrow p$
(b). $p \Rightarrow q$
(c). $\sim q \Rightarrow \sim p$
(d). $\sim p \Rightarrow \sim q$
Show Answer
Answer: (a)7. $\sim(p \vee q) \vee(\sim p \wedge q)$ is equivalent to
(a). $\mathrm{q}$
(b). $\mathrm{p}$
(c). $\sim \mathrm{p}$
(d). $\sim \mathrm{q}$
Show Answer
Answer: (c)8. Negation of the compound proposition. If the examination is difficult, then I shall pass if I study hard.
(a). The examination is difficult and I study hard but I shall not pass
(b). The examination is difficult and I study hard and I shall pass
(c). The examination is not difficult and I study hard and I shall pass
(d). None of these
Show Answer
Answer: (a)9. If $p$ is true, $q$ is false and $r$ is false, then which of the following is true?
(a). $(p \vee q) \Rightarrow r$
(b). $\mathrm{p} _{\wedge} \sim(\mathrm{q} \vee \mathrm{r})$
(c). $(p \vee q) \wedge r$
(d). $\mathrm{p} \Rightarrow \sim(\mathrm{q} \Rightarrow \mathrm{r})$
Show Answer
Answer: (b)10. If $\mathrm{p} _{\equiv}$ “she goes to market” and $\mathrm{q} \equiv$ “She buys some fruits”.
Then choose the correct symbol for the given statements :
(i). Either she goes to market or she buys some fruits:
(a). $\mathrm{p} \vee \mathrm{q}$
(b). $\mathrm{p} \wedge \mathrm{q}$
(c). $ \sim \mathrm{p} \vee \mathrm{q}$
(d). $\mathrm{p} _{\vee} \sim \mathrm{q}$
Show Answer
Answer: (a)(ii). If she goes to market, then she buys some fruits:
(a). $\sim \mathrm{p} \wedge \mathrm{q}$
(b). $\mathrm{p} \Rightarrow \mathrm{q}$
(c). $\mathrm{p} \wedge \mathrm{q}$
(d). $\sim p _{\vee} q$
Show Answer
Answer: (b)(iii). Neither she go to market nor she buy some fruits:
(a). $\sim \mathrm{p} \vee \sim \mathrm{q} $
(b). $\mathrm{p} \Rightarrow \mathrm{q}$
(c). $\mathrm{p} \vee \mathrm{q} $
(d). $\sim p _{\wedge} \sim q$
Show Answer
Answer: (d)(iv). She does not go to market and she buys some fruits:
(a). $\sim \mathrm{p} \wedge \mathrm{q}$
(b). $\sim(p \vee q)$
(c). $\mathrm{p} _{\wedge} \sim \mathrm{q}$
(d). $p \Rightarrow q$
Show Answer
Answer: (a)(v). She does not go to market unless she buys some fruits :
(a). $\mathrm{p} \Rightarrow \mathrm{q}$
(b). $q \Rightarrow p$
(c). $\sim q \Rightarrow \sim p$
(d). $\mathrm{p} _{\vee} \mathrm{q}$
