ACE Test series: Propositional Logic

Question

Which of the following statement is valid?

$A)\left \{ \forall \left \{ P\left ( x \right )\vee Q\left ( x \right ) \right \}, \forall x Q\left ( x \right ) \right \}\Rightarrow \exists x P\left ( x \right )$

$B) \left \{ \forall x P\left ( x \right ) , \exists x \sim \left \{ P\left ( x \right ) \wedge Q\left ( x \right )\right \} \right \} \Rightarrow \exists x \sim Q\left ( x \right )$

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 We cannot solve it putting simple logic,  We need to use propositional formula here, right?

Answer 1

$ \forall \left \{ P\left ( x \right )\vee Q\left ( x \right ) \right \}$  //Either P or Q should be true.

      $ \forall x Q\left ( x \right )$   //Q is always true.

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$\therefore  \exists x P\left ( x \right )$ // This means Q(x) is always True so P(x) can be true or false. So it is possible that $ \exists x P\left ( x \right )$ is True but we can't say surely about it.

so $A)\left \{ \forall \left \{ P\left ( x \right )\vee Q\left ( x \right ) \right \}, \forall x Q\left ( x \right ) \right \}\Rightarrow \exists x P\left ( x \right )$ is not valid but it is satisfiable.

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$ \forall x P\left ( x \right ) $ //P is always true

$ \exists x \sim \left \{ P\left ( x \right ) \wedge Q\left ( x \right )\right \} $ \\ There exist a value of $x$ such that P and Q are not true at same time.

But we know P is always True. hence Q must be false for atleast one value of $x$.

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$\therefore  \exists x \sim Q\left ( x \right )$ \\ There exist a value of $x$ for which Q is not true.

so $B) \left \{ \forall x P\left ( x \right ) , \exists x \sim \left \{ P\left ( x \right ) \wedge Q\left ( x \right )\right \} \right \} \Rightarrow \exists x \sim Q\left ( x \right )$ is valid

Comments

$\therefore  \exists x \sim Q\left ( x \right ) \equiv \sim \forall x  Q\left ( x \right ) $ 

I couldnot understand, how this line could be correct?

For some $x$ $Q(x)$ doesnot exists, does it mean ,"For all x Q is false or not exists" ?? 

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It means " there  exist some value of x for which Q is not true"

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RHS u are using for all

Then how

It means " there  exist some value of x for which Q is not true"

this statement could be true?

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I thought in reverse.

what would be negation of $\forall x Q(x)$ ? i.e. $\sim \forall x Q(x)$ ?

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Ok i got it.

LHS

= there  exist some value of x for which Q is not true

i.e. for some value of x Q is true and for some value it is false

i.e. for all values of x Q(x) is not true

=RHS

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Moreover check this point too

In A) the sign is $\Rightarrow$ and not $\Leftrightarrow$

Then say $\exists x P\left ( x \right )$ could be NULL set too, tha means there exists nothing.

So, why A) is not true??

Take ur time and think this point plz.

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$\exists P(x)$ means there exist atleast one value of $x$ for which P(x) is true.

so $\exists P(x)$ can't  be null set.

As i already mentioned from the given premises we can't derive the conclusion always. So A is false.

  it is possible that ∃xP(x) is True but we can't say surely about it.

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Second statement meaning and derivation both are wrong @Satbir

but the result is correct

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Ok i understood. I can't pull out negation like that. Now corrected please check.

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" Q must be false for atleast one case " but not all