$\text{BF(x,y)} = \text{x and y are best friends}$

$S1 : \forall x \exists y \forall z (BF(x,y) \wedge \neg BF(x,z) \implies (y \neq z))$

This statement says

for all $x$ there exist a $y$ and for all $z$

if $(x$ and $y$ are best friends and if $x$ and $z$ are not best friends $)$ then this means that $y$ and $z$ are not same people.

which is trivially true.(how can $y$ be both best friend and **NOT **best friend of $x$ at same time)

So this statement is not equivalent to “Everyone has exactly one best friend”

$S2: \forall x \exists y (BF(x,y) \implies \forall z [(y \neq z)] \implies \neg BF(x,z)) $

We can rewrite this as

$S2: \forall x \exists y (BF(x,y) \wedge \forall z [(y \neq z)] \implies \neg BF(x,z)) $ (since implication has right asociatiivity)

This statement says

for all $x$ there exist a $y$ such that

if $x$ and $y$ are best friends

AND

among all $z$ whichever $z$ we select is not equal to $y$ i.e. $y$ and $z$ are different people

then $x$ and $z$ can’t be best friends.

This statement is all not equivalent to “Everyone has exactly one best friend” because if for any $x$ there isn’t any best friend then also $S2$ becomes true because $F⟹anything \equiv True \vee anything \equiv True$

Hence both $S1$ and $S2$ are not correct.