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

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.