Let $p, q, r, s$ represents the following propositions.
The integer $x\geq2$ which satisfies $\neg\left(\left(p\Rightarrow q\right) \wedge \left(\neg r \vee \neg s\right)\right)$ is ____________.
$\neg((p → q) \wedge (\neg r \vee \neg s))$ $\quad \equiv (\neg(\neg p \vee q)) \vee (\neg(\neg r \vee \neg s))$ $\quad \equiv (p \wedge \neg q) \vee (r \wedge s)$
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
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 $11.$
~((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
arush_verma
we can replace ^ with Ո and ˅ with U.
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