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My doubt is why D) can't be the answer

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CAN anyone give me explanation for it , i think d) can also be the answer

Either Or means $\text{XOR}$

Either everything is material or there are somethings that are not material

we'll define p as $\forall x (\text{material}(x))$

we'll define q as $\exists x (¬\text{material}( x))$

Either p or q $\equiv$ $(¬p \wedge q) \vee (p \wedge ¬q)$

$(¬ \forall x (\text{material}(x)) \wedge \exists x (¬\text{material}( x))) \vee ( \forall x (\text{material}(x)) \wedge ¬\exists x (¬\text{material}( x)))$

$(\exists x (¬\text{material}( x))\wedge \exists x (¬\text{material}( x))) \vee ( \forall x (\text{material}(x)) \wedge \forall x (\text{material}(x)) )$

$(\exists x (¬\text{material}( x)) \vee ( \forall x (\text{material}(x)) )$

Hence Option A is correct

Now the question is why D is incorrect?

$\forall x (M(x)) \vee \exists y (¬ M(x))$

for example $X = \{ 1 , 2 , 3 , 4, 5 \}$ and $M = \{2,3,4 \}$ and $Y = \{ 2,4 \}$

So by this example we can see that $\forall x (M(x))$ is False and also $\exists y (¬ M(x))$ is false because both 2 and 4 are Materials.

So overall $F \vee F$, but this conclusion is incorrect. So D is incorrect.

I think option d ) should also true because scope of both quantifier is different so why it iwill create any problem

Prince Sindhiya i just gave you the example when problem was created

it is ∃y(¬M(y)) not

yes $\exists y(¬M(y))$ which means the expression is true when there exists a y which is not a material by Y consists of elements which are all in M (all are material).

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