Tag: discrete mathematics
Questions Related to discrete mathematics
Which of the following statement is a contradiction ?
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$(\sim p \vee \sim q) \vee (p \vee \sim q)$
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$(p \rightarrow q) \vee (p \wedge \sim q)$
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$(\sim p \wedge q) \wedge (\sim q)$
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$(\sim p \wedge q) \vee (\sim q)$
If $p$ is any statement, $t$ is a tautology and $c$ is a contradiction, then which for the following is NOT correct?
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$p \wedge (\sim c) \equiv p$
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$p \vee (\sim t) \equiv p$
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$t \vee c \equiv p \vee t$
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$(p\wedge t) \vee (p \vee c) \equiv (t \wedge c)$
A. $p \wedge (\sim c )\equiv p \wedge t \equiv p$
If $p$ is any statement, $t$ and $c$ are a tautology and a contradiction respectively, then which of the following is INCORRECT?
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$p \wedge t \equiv p $
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$ p \wedge c \equiv c$
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$p \vee t \equiv p $
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$ p \vee c \equiv p$
Truth table,
| $p$ | $p\wedge t$ | $p\wedge c$ | $p\vee t$ | $p\vee c$ |
|---|---|---|---|---|
| T | T | F | T | T |
| F | F | F | T | F |
Hence option 'C' is the correct choice.
The statement $(p \rightarrow ~p) \wedge (~ p \rightarrow p)$ is
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a tautology.
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a contradiction.
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neither a tautology nor a contradiction.
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None of these.
If $p$ is a true statement, then $p \rightarrow p$ is true.
Which of the following statement is a contradiction?
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$(p \wedge q) \wedge (\sim(p \vee q))$
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$p \vee (\sim p \wedge q)$
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$(p \rightarrow q) \rightarrow p$
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$\sim p \vee \sim q$
We check for contradiction for all the given options.
| $p$ | $q$ | $\left( p\wedge q \right)$ | $\left( p\vee q \right)$ | $\left( \sim \left( p\vee q \right) \right)$ | $\left( p\wedge q \right) \wedge \left( \sim \left( p\vee q \right) \right) $ |
|---|---|---|---|---|---|
| T | T | T | T | F | F |
| T | F | F | T | F | F |
| F | T | F | T | F | F |
| F | F | F | F | T | F |
All F so this is a contradiction.
| $p$ | $q$ | $\sim p$ | $\sim p\wedge q$ | $p\vee \left( \sim p\wedge q \right) $ |
|---|---|---|---|---|
| T | T | F | F | T |
| T | F | F | F | T |
| F | T | T | T | T |
| F | F | T | F | F |
So not a contradiction.
| $p$ | $q$ | $\left( p\longrightarrow q \right) $ | $\left( p\longrightarrow q \right) \rightarrow p$ |
|---|---|---|---|
| T | T | T | T |
| T | F | F | T |
| F | T | T | F |
| F | F | T | F |
So it is also not a contradiction.
| $p$ | $q$ | $\sim p$ | $\sim q$ | $\sim p\vee \sim q$ |
|---|---|---|---|---|
| T | T | F | F | F |
| T | F | F | T | T |
| F | T | T | T | T |
| F | F | T | T | T |
The statement $\sim (p \rightarrow q )\leftrightarrow (\sim p \vee \sim q)$ is
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a tautology.
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a contradiction.
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neither a tautology nor a contradiction.
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None of these.
When $p$ and $q$ both are true then
Which of the following is a tautology?
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$p\implies p\wedge q$
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$p\implies p\vee q$
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$(p\vee q)\implies(p\wedge q)$
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None of these
A tautology is a statement that is always true.
The statement $ p\Longrightarrow p\vee q $ is read as " if p is true, then either p or q is true. " as the symbol $ \vee $ denotes OR.
Hence, the given statement is a tautology.
Which of the following statements is/are true?
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$p\wedge (\sim p)$ is a contradiction.
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$(p\rightarrow q)\Leftrightarrow (\sim q \rightarrow \sim p)$ is a contradiction.
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$\sim(\sim p) \Leftrightarrow p$ is a tautology.
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$p\vee (\sim p)$ is a tautology.
A,C,D obvious and for B
p $\rightarrow$ q is same as $\sim$ q $\rightarrow$ $\sim$p
$\therefore$ it is tautology not contradiction.
If $p$ is any statement $t$ and $c$ are tautology and contradiction respectively, then which of the following is(are) correct?
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$p\wedge t \equiv p$
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$p \wedge c \equiv c$
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$p\vee t \equiv c$
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$p \vee c \equiv p$
Tautology is the preposition which is always true and contradiction is a preposition which is always false.
Here $\wedge \equiv AND, \vee \equiv OR$, $T$=true and $F$=false
Given statement=$p$, tautology=$t$ and contradiction=$c$
a)$p\wedge t\equiv p$ irrespective of value of $p$ as $t=T$ always
And if $p=T$ then $T\wedge T=T$ and if $p=F$ then $F\wedge T=F$
Thus option (a) is correct.
b)$p\wedge c\equiv c$ irrespective of value of $p$ as$ c=F$ always
And if $p=T$ then $T\wedge F=F$ and if $p=F$ then $F\wedge F=F$
Thus option (b) is correct
c)$p\vee t\equiv c$
If $p=T$ then $T\vee T\equiv T\neq c$ and if $p=F$ then $F\vee T\equiv T \neq c$
Thus option (c) is not correct
d)$p\vee c\equiv p$
If $p=T$ then $T\vee F\equiv T$ and if $p=F$ then $F\vee F\equiv F$
Thus it depends on the value of $p$
Hence option (d) is correct
Which one of the following statements is a tautology?
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$\left( p\vee q \right) \rightarrow q$
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$p\vee (p\rightarrow q)$
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$ p\vee (q\rightarrow p)$
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$p\rightarrow (p\rightarrow q)$
| $ p$ | $q$ | $p\to q$ | $q\to p$ | $pvq$ | $p\to (p\to q)$ | $(pvq)\to q$ | $pv(p\to q)$ | $pv(q\to q$ |
|---|---|---|---|---|---|---|---|---|
| T | T | T | T | T | T | T | T | T |
| T | F | F | T | T | F | F | T | T |
| F | T | T | F | T | T | T | T | F |
| F | F | T | T | F | T | T | T | T |
$\left( p\vee q \right) \rightarrow q$