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Everyone has exactly one best friend

Which of the following first order logic statements correctly represents above English statement? $BF(x,y)$  means $x$ and $y$ are best friends

• $S1 : \forall x \exists y \forall z (BF(x,y) \wedge \sim BF(x,z) \rightarrow (y \neq z))$
• $S2 : \forall x \exists y (BF(x,y)\rightarrow \forall z [(y \neq z)] \rightarrow \sim BF(x,z))$
1. Only $S1$
2. Only $S2$
3. Both $S1$ and $S2$
4. None of the two
edited | 137 views
0

@Soumya29 PLease guide with this

2nd : False

if x does not have any best friend (y), then statement is becoming false and overall giving TRue., same reason with 1st option

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@Learner_jai answered. Check it,

+1

Answer- Option $D$
$\text{Everyone has exactly one best friend}$

We can express it as -

$\forall x \exists y(BF(x,y) \wedge \forall z(BF(x,z) \rightarrow (y = z)))$

$\text{OR equivalently we can write it as - }$

$\forall x \exists y(BF(x,y) \wedge \forall z( (y \neq z)\rightarrow \sim BF(x,z)))$

$S1 -$ It is same as -$\forall x \exists y \forall z ((BF(x,y) \wedge \sim BF(x,z) )\rightarrow (y \neq z))​​​​​​$
It is false because if some person has no best friend then LHS will be false so overall statement becomes true.

$S2-$ is false. Same reasoning as above.
answered by Boss (15.3k points)
0

∀x∃y(BF(x,y)∧∀z(BF(x,z)→(y=z)))

Here in this equation which has highest precedence is it    $\wedge$  or  $\rightarrow$. Lets say there is no bracket defined then which one should be considered.

As given in book the precedence of operators is this

~ , $\wedge , \vee , \Rightarrow , \Leftrightarrow$

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@nagendra2016
I removed brackets here - $\forall x \exists y(BF(x,y) \wedge \forall z \ BF(x,y) \rightarrow (y=z))$
Now, $\forall$ is applicable only on $BF(x,z).$
And  will take this 2 -$BF(x,y) \wedge \forall z \ BF(x,y)$