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71 votes
71 votes

Let $p, q, r, s$ represents the following propositions.

  • $p:x\in\left\{8, 9, 10, 11, 12\right\}$
  • $q:$ $x$ is a composite number.
  • $r:$ $x$ is a perfect square.
  • $s:$ $x$ is a prime number.


The integer $x\geq2$ which satisfies $\neg\left(\left(p\Rightarrow q\right) \wedge \left(\neg r \vee \neg s\right)\right)$ is ____________.

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7 Answers

Best answer
130 votes
130 votes

$\neg((p → q) \wedge (\neg r \vee \neg s))$
$\quad \equiv (\neg(\neg p \vee q)) \vee (\neg(\neg r \vee \neg s))$
$\quad \equiv (p \wedge \neg q) \vee (r \wedge s)$

which can be read as $(x\in  \{8,9,10,11,12\}$ AND $x$ is not a composite number$)$ OR $(x$ is a perfect square AND $x$ is a prime number$)$
Now for

  • $x$ is a perfect square and $x$ is a prime number, this can never be true as every square has at least $3$ factors, $1,x$ and $x^2.$

So, second condition can never be true.
which implies the first condition must be true.
$x\in \{8,9,10,11,12\}$ AND $x$ is not a composite number

But here only 11 is not a composite number. so only $11$ satisfies the above statement. 

ANSWER $11.$

edited by
8 votes
8 votes

~((p → q) ⋀ (~r ⋁ ~s)) 
= (~(~p ⋁ q)) ⋁ (~(~r ⋁ ~s)) 
=(p ⋀ ~q) ⋁ (r ⋀ s)

now all whole numbers > 1 are either prime or composite . 

taking that into consideration..  ~q is equal to s

so equation becomes (p ⋀ ~q) ⋁ (r ⋀ s) = (p ⋀ s) ⋁ (r ⋀ s) = ( p ⋁ r ) ⋀ s 

now to above expression to become tautology .. s should be true...

only prime number in our set is 11...

So answer would be 11

7 votes
7 votes
(p ⇒ q) will give {8, 9, 10, 12} ¬r will give {8, 10, 11, 12} ¬s will give {8, 9, 10, 12} (¬r ∨ ¬s) will give {8, 9, 10, 11, 12} (p ⇒ q) ∧ (¬r ∨ ¬s) will give {8, 9, 10, 12} ¬((p ⇒ q) ∧ (¬r ∨ ¬s)) will give 11. Thus, C is the correct option.
1 votes
1 votes
Answer 11
Answer:

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