Tag: different forms of theoretical statements
Questions Related to different forms of theoretical statements
The dual of the statement $\sim p \wedge [\sim q \wedge (p \vee q) \wedge \sim r]$ is:
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$\sim p \vee [\sim q \vee (p \vee q) \vee \sim r]$
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$ p \vee [q \vee (\sim p \wedge \sim q) \vee r]$
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$ \sim p \vee [\sim q \vee (p\wedge q) \vee \sim r]$
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$ \sim p \vee [\sim q \wedge (p\wedge q) \wedge \sim r]$
The dual of the statement $\sim p \wedge [\sim q \wedge (p \vee q) \wedge \sim r]$ is
Which of the following is equivalent to $(p \wedge q)$?
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$p \rightarrow \sim q $
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$ \sim (\sim p \wedge \sim q)$
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$ \sim ( p \rightarrow \sim q)$
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None of these
$p \wedge q \equiv \sim (\sim p \vee \sim q) \equiv \sim (p \to \sim q)$
Which of the following is equivalent to $( p \wedge q)$?
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$p \rightarrow \sim q$
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$\sim (\sim p \wedge \sim q)$
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$\sim (p \rightarrow \sim q)$
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None of these.
$p \wedge q \equiv \sim (\sim p \vee \sim q) \equiv \sim (p \to \sim q)$
The equivalent statement of (p $\leftrightarrow$ q) is
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$(p \wedge q) \vee (p \vee q)$
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$(p \rightarrow q) \vee (q \rightarrow p)$
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$(\sim p \vee q) \vee (p \vee \sim q)$
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$(\sim p \vee q) \wedge (p \vee \sim q)$
$p\rightarrow q \equiv (\sim p\vee q)$
Which of the following is correct?
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$(~p \vee ~q) \equiv (p \wedge q)$
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$(p \rightarrow q) \equiv (~q \rightarrow ~p)$
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$~(p \rightarrow ~q) \equiv (p \wedge ~q)$
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none of these
Clearly, the statements $p \vee q$ and $p\wedge q$ cannot be equivalent as they one operator means "OR" and the other operator means "AND".
| $p$ | $q$ | $p\rightarrow q$ | $q\rightarrow p$ |
|---|---|---|---|
| T | T | T | T |
| T | F | F | T |
| F | T | T | F |
| F | F | T | T |
Option B is also incorrect.
| $p$ | $q$ | $p\rightarrow q$ | $p\wedge q$ |
|---|---|---|---|
| T | T | T | T |
| T | F | F | F |
| F | T | T | F |
| F | F | T | F |
Hence, option C is also incorrect.
Option D is also incorrect as
$p \leftrightarrow q=(p\rightarrow q)\wedge (q\rightarrow p)$
Which of the following statement are NOT logically equivalent?
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$ \sim (p \vee \sim q)$ and $ (\sim p \wedge q )$
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$\sim (p \rightarrow q )$ and $(p \wedge \sim q )$
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$(p \rightarrow q) $ and $(\sim q \rightarrow \sim p) $
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$(p \rightarrow q )$ and $(\sim p \wedge q)$
We make an option wise check for this.
$(~ p \vee ~ q)$ is logically equivalent to
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$(p \wedge q) \vee (p \vee q)$
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$(p \rightarrow q) \vee (q \rightarrow p)$
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$(\sim p \vee q) \vee (p \vee \sim q)$
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$(\sim p \vee q) \wedge (p \vee \sim q)$
The statement $\sim (p\rightarrow \sim q)$ is equivalence to ___________.
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$(\sim p\vee q)$
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$(p\vee \sim q)$
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$(\sim p\wedge q)$
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$(p\wedge \sim q)$
$\sim\left({p} \rightarrow \sim{q} \right)$
Which of the following is always true?
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$\sim(p\rightarrow q) \equiv \sim p \wedge q$
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$\sim(p\vee q) \equiv \sim p \vee \sim q$
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$\sim (p \implies q ) \equiv (p \land \sim q )$
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$\sim(p \wedge q) \equiv \sim p \wedge \sim q$
$p \implies q \equiv \sim p \lor q $
$\therefore \sim (p \implies q ) \equiv \sim (\sim p \lor q )$
$\therefore \sim (p \implies q ) \equiv (p \land \sim q )$
Which of the following is/are false?
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$p\rightarrow q\equiv\sim p\rightarrow\sim q$
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$\sim(p \rightarrow\sim q)\equiv\sim p\wedge q$
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$\sim(\sim p\rightarrow\sim q)\equiv\sim p\wedge q$
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$\sim (p\leftrightarrow q) \equiv(\sim(p\rightarrow q))\wedge\sim(q\rightarrow p)$
We know that:
$p\rightarrow q \equiv \sim q\rightarrow \sim p$ {By logical equivalences }
Hence $A$ is false
$\sim (p \ \rightarrow \ \sim q)$ $\equiv$ $\sim (\sim p\vee \sim q)=p\wedge q$ [By logical Equivalences ]
Hence $B$ is false
$\sim (\sim p\rightarrow \sim q)$ $\equiv \sim (p \vee \sim q) $ $\equiv \sim p\wedge q$ [By Logical Equivalences]
Hence $C$ is true
$\sim (p\leftrightarrow q)$ $\equiv \sim ((p\rightarrow q)\wedge (q\rightarrow p))$ $\equiv \sim (p\rightarrow q)\vee \sim (q\rightarrow p)$
Hence $D$ is false [By logical Equivalences]