Self doubt- Propositional Logic

Question

Que. Consider domain is the set of all people in the world.

$F(x,y) =x \text{ is the friend of y}.$ 

Represent each of the following sentences using first-order logic statements
$1.$ Every person has $at most \ 2$ friends.
$2.$ Every person has $exactly \ 2$ friends.
$3.$ Every person has $at least \ 2$ friends. 
 

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My attempt –
$1. \forall x \exists y_1\exists y_2(F(x,y_1) \wedge F(x,y_2) \wedge \forall z(F(x,z) \implies ((z= y_1) \vee (z= y_2))))$
$2. \forall x \exists y_1\exists y_2(F(x,y_1) \wedge F(x,y_2) \wedge (y_1 \neq y_2) \wedge  \forall z(F(x,z) \implies ((z= y_1) \vee (z= y_2))))$
$3. \forall x \exists y_1\exists y_2(F(x,y_1) \wedge F(x,y_2) \wedge (y_1 \neq y_2))$

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Please verify. 

 

Comments

https://gateoverflow.in/37967/made-easy

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Every person has $atmost$ 2 friends. - means he can have either 0, 1 or 2 friends. According to me it should be

$\forall x \exists y_1\exists y_2(F(x,y_1) \wedge F(x,y_2) \wedge \forall z(F(x,z) \implies (((z \neq y_1) \wedge (z \neq y_2))\vee (z= y_1) \vee (z= y_2)))))$

Let me know if i am making any mistake in this.

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@!KARAN, I don't think so it will work. Try when x has 3 friends. 
There is a flaw in my formula too. It's not covering "0" case, 

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@Soumya29

For Statement 1, If you use "for all" symbol then it should work.

 $\forall x \forall y_1 \forall y_2(F(x,y_1) \wedge F(x,y_2) \wedge \forall z(F(x,z) \implies ((z= y_1) \vee (z= y_2))))$

or simply write- 

 $\forall x \forall y_1 \forall y_2 \forall z (F(x,y_1) \wedge F(x,y_2) \wedge F(x,z) \implies ((z= y_1) \vee (z= y_2)))$

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Yes, it covers all the cases now. Thank you so much @Sachin   :)

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@Deepak Poonia sir 

correct me If i am wrong 

in the above formula we have to include the condition (y1=y2) becoz it possible for me to have case like this 

when both y1=y2 and for z which not both y1 and y2 then the implication part results in false then whole stmt results in false but here I just have only 2frds for x 

so that's why we have to include the condition (y1=y2). Correct me If I am wrong sir.

 

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@Lakshmi Narayana404

Yes, you are right. We have seen “At most two” Quantification in this lecture, in detail.

Correct expression will be:

For Statement 1:

 $\forall x \forall y_1 \forall y_2 \forall z (F(x,y_1) \wedge F(x,y_2) \wedge F(x,z) \implies ((z= y_1) \vee (y_1 = y_2) \vee (z= y_2)))$

We have seen ALL these Numerical Quantification in this playlist, in detail.