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Consider the following formula and its two interpretations $I_1$ and $I_2$.

$\alpha: (\forall x)\left[P_x \Leftrightarrow (\forall y)\left[Q_{xy} \Leftrightarrow \neg Q_{yy} \right]\right] \Rightarrow (\forall x)\left[\neg P_x\right]$

$I_1$ : Domain: the set of natural numbers

$P_x$ = 'x is a prime number'

$Q_{xy}$ = 'y divides x'

$I_2$ : same as $I_1$ except that $P_x$ = 'x is a composite number'.

Which of the following statements is true?

1. $I_1$ satisfies $\alpha$, $I_2$ does not
2. $I_2$ satisfies $\alpha$, $I_1$ does not
3. Neither $I_1$ nor $I_2$ satisfies $\alpha$
4. Both $I_1$ and $I_2$ satisfies $\alpha$
edited | 2.4k views
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What does "satisfies" mean here?? (Does it mean "satisfiability"?)

Are the terms "satisfiability" and "valid" equivalent??
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satisfiability means true for at least one instance, valid means always true(true for all instances).

Given: $(\alpha: (\forall x)\left[P_x \Leftrightarrow (\forall y)\left[Q_{xy} \Leftrightarrow \neg Q_{yy} \right]\right] \Rightarrow (\forall x)\left[\neg P_x\right])\\$
$Q_{yy}$ is always True, this makes  $\neg Q_{yy}$ False.

Writing $(\forall y)[Q_{xy} \Leftrightarrow \neg Q_{yy}]$ is same as writing $(\forall y)[Q_{xy} \Leftrightarrow False]]$

This is equivalent to saying that, for all y $Q_{xy}$ is false and finally we can rewrite $(\forall y)[Q_{xy} \Leftrightarrow \neg Q_{yy}]]$ as $(\forall y)[\neg Q_{xy}]$

α: (∀x)[Px⇔(∀y)[¬Qxy]]⇒(∀x)[¬Px

LHS:(∀x)[Px⇔(∀y)[¬Qxy]]

consider only (∀y)[¬Qxy] it says all values of y does not divide x, but there will be atleast one value of y (when y=x, or when y=1) that divides x, i.e. [¬Qxy] is not true for all values of y.  (∀y)[¬Qxyis false.

Now LHS becomes (∀x)[Px⇔False], "Px⇔False" this means Pis False, which is same as writing "¬Px".

Finally, we reduced LHS to (∀x)[¬Px]

α: (∀x)[¬Px]⇒(∀x)[¬Px]  (which is trivial, P(x)⇒ P(x) is trivially true)

Hence (∀x)[¬Px]⇒(∀x)[¬Px]  is trivially true for any P(x), doent matter if P(x) is for prime number or for composite number (I1 or I2).

⇒ I1 and I2 both satisfies α.

Option D.

Thanks Soumya29 for suggesting edit.

edited
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Awesome explanation @sachin Mittal  (Y)
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thnks :)
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Correct and clear explanation sachin..
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Can you please explain how we can rewrite

(∀y)[Qxy⇔¬Qyy]](∀y)[Qxy⇔¬Qyy]] as (∀y)[¬Qxy]?
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I have doubt in this solution.As mentioned

"consider only (∀y)[¬Qxy] it says all values of y does not divide x, It is true irrespective of x being prime or composite. Even if x is composite then only its factor divides x not all values."

How is this true always? when y and x are same in case of prime/non prime  then it will divide.

(∀y)[¬Qxy] says that there is no number that divides x and so it should be false because if there is one case which is not satisfying for all then it is not FOR ALL.Please correct me if i understood it wrong.

Please tell me if following approach will work:-

α:(∀x)[Px⇔(∀y)[Qxy⇔¬Qyy]]⇒(∀x)[¬Px]

We know y will always divide y(assuming 0 is not a natural number as in group theory it says 0 is natural number set but some says if 0 is included then it becomes whole number.)

A Qyy is always false,means Qxy=False.

So i can reduce this to:-

(∀x)[Px⇔(∀y)[False]]

which means

(∀x)[Px= false]

(∀x)[false]

and false implies anything must be true?

Please tell me if i am wrong in this approach?

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sir u r really talented
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Nice explanation.Thanks,  Sachin Mittal 1.

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## Bi conditional Truth table ...

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Just for thanking you I created account

Whenever I encounter your answer I can be assure that it must be checked or from trusted reference.

Thanks @sachin for your great effort and indeed wonderful explanation :)
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@Jordan Thank you so much. :)

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@Sachin

(∀y)[¬Qxy] can also be interpreted as -
¬∃(y)[Qxy] - There doesn't exist a number y such that y divides x.
Which is false always as 1 divides every number - Counter-example
So LHS:(∀x)[Px⇔(∀y)[¬Qxy]] reduced to (∀x)[Px⇔False] .
I can write it as (∀x)[¬Px
(∀x)[¬Px] ⇒ (∀x)[¬Px]
F⇒F
T

Correct me if I am wrong.  :)

$\alpha: (\forall x)\left[P_x \Leftrightarrow (\forall y)\left[Q_{xy} \Leftrightarrow \neg Q_{yy} \right]\right] \Rightarrow (\forall x)\left[\neg P_x\right]$

This is can be interpreted as:

• $\alpha: \left( (\forall x)\left[P_x \Leftrightarrow (\forall y)\left[Q_{xy} \Leftrightarrow \neg Q_{yy} \right]\right] \right) \Rightarrow \left((\forall x)\left[\neg P_x\right] \right)$

See the RHS. It says $P(x)$ is false for any natural number. But there are natural numbers which are prime and hence this RHS is FALSE. Now, to make $\alpha$ TRUE, LHS must be FALSE for any $x$. Here, LHS is bit complex, so lets consider it separately.

$(\forall x)\left[P_x \Leftrightarrow (\forall y)\left[Q_{xy} \Leftrightarrow \neg Q_{yy} \right]\right]$

LHS is TRUE only if the given implication is TRUE for all $x$. Here the rightmost double implication $(\forall y)\left[Q_{xy} \Leftrightarrow \neg Q_{yy} \right]$ is always FALSE, because $x$ can be equal to $y$ and hence forall can never be TRUE. So the LHS reduces to just $(\forall x) \neg P(x)$ and returns FALSE as we have prime as well as non-prime natural numbers. So, FALSE $\Rightarrow$ FALSE returns TRUE making both $I_1$ and $I_2$ satisfy $\alpha$. D choice.

edited by
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option given by ,
GK publication - (A) ,
Ankur Gupta website - (C) ,
and happy mittal sir - (C) , http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2003.html
@ Arjun sir , I think you are right(B) , but I am still doubt , please remove my doubt , with the help of a example .
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Sorry, my answer was wrong. I had missed the possibility of $x = y$. Also, two interpretations are possible here as shown now.
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@Happy-Mittal These are the two cases for 'precedence'.
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The scope of quantified variable is determined by square brackets, which is clear from the question and hence your second interpretation is correct.
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Also, your first interpretation doesn't make sense, as once we are arguing about all x, then we don't write $\forall x$ inside that, we are anyway arguing about every x.
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That make sense. I'll change. Thanks :)
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@Arjun, So the LHS reduces to just ~P(X), make this change for more correctness & Readablity !
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Can you verify best answer selected?

In that LHS: (∀x)[Px].But LHS should be reduced to LHS: (∀x)[¬Px].

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@rahul sharma 5 (∀x)[Px<->TRUE] reduced to (∀x)[Px] because final value depend on Px only,expand and see yourself !

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(∀x)[Px⇔(∀y)[Qxy⇔¬Qyy]]

Now [Qxy⇔¬Qyy] becomes false as mentioned in answer.

(∀x)[Px⇔(∀y)[False]]

(∀x)[Px⇔[False]]

(∀x)[¬Px]

Please tell if i am wrong
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@rahul sharma 5 correct !