Tag: discrete mathematics
Questions Related to discrete mathematics
$(p \wedge \sim q)\wedge (\sim p \vee q)$ is
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tautology
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contradiction
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dualoty
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double implication
Which of the following is not correct ?
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$p \vee \sim p $ is a tautology.
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$\sim (\sim p) \leftrightarrow p$ is a tautology.
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$p \wedge \sim p $ is a contradiction.
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$([(p \wedge p ) \rightarrow q] \rightarrow p)$ is a tautology.
$\because [(p \wedge p ) \rightarrow q ] \rightarrow p \equiv ( p
\rightarrow q ) \rightarrow p (\because p \wedge p \equiv p)$
Which of the following statement is a tautology?
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$(\sim p \vee \sim q ) \vee ( p \vee \sim q )$
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$(\sim p \vee \sim q ) \wedge (p \vee \sim q )$
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$\sim p \wedge (\sim p \vee \sim q )$
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$\sim q \wedge (\sim p \vee \sim q )$
$\because (\sim p \vee \sim q) \vee (p \vee \sim q) $
$ p\Rightarrow p \vee q$ is
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a tautology.
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a contradiction.
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a tautology and a contradiction.
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neither a tautology nor a contradiction.
$p$ | $q$ | $p\vee q$ | $p\rightarrow (p\vee q)$ |
---|---|---|---|
T | T | T | T |
T | F | T | T |
F | T | T | T |
F | F | F | T |
Since, all the entries in the last column has true value. So, the given statement is a tautology
$p\Rightarrow \sim p$ is
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a tautology.
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a contradiction.
-
a tautology and a contradiction.
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neither a tautology nor a contradiction.
$p$ | $\sim p$ | $p\Rightarrow \sim p$ |
---|---|---|
T | F | F |
F | T | T |
So the result of the Truth table show that $p\Rightarrow \sim p$ is neither a tautology not a contradiction
$ p\wedge (\sim p)$ is
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a tautology.
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a contradiction.
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a tautology and a contradiction.
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neither a tautology nor a contradiction.
Truth table is,
$p$ | $\sim p$ | $p\wedge (\sim p)$ |
---|---|---|
T | F | F |
F | T | F |
Hence given statement is contradiction.
Which of the following is a contradiction?
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$p\vee q$
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$p\wedge q$
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$p\vee (\sim p)$
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$p\wedge (\sim p)$
$p$ | $q$ | $p\vee q$ | $p\wedge q$ |
---|---|---|---|
$ T$ | $ T$ | $ T$ | $ T$ |
$ T$ | $ F$ | $ T$ | $ F$ |
$ F$ | $T$ | $T$ | $F$ |
$ F$ | $F$ | $ F$ | $ F$ |
The proposition $(p\rightarrow \sim p)\wedge (\sim p\rightarrow q)$ 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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a tautology and a contradiction
$p$ | $q$ | $\sim p$ | $p\rightarrow \sim p$ | $\sim p \rightarrow q$ | $(p\rightarrow \sim p)\wedge(\sim p\rightarrow q)$ |
---|---|---|---|---|---|
T | T | F | F | T | F |
T | F | F | F | T | F |
F | T | T | T | T | T |
F | F | T | T | F | F |
The statement $(p-q)\rightarrow [(\sim p \rightarrow q)\rightarrow q]$ is
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a tautology
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equivalent to $\sim p \rightarrow q$
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equivalent to $p\rightarrow \sim q$
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a fallacy
Which of the following proposition 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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$(\sim p\wedge q)\vee (\sim q)$
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