Assuming that ⊢ is sound and complete with respect to ⊨, select all statements that are valid For this question:

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*Assuming that ⊢ is sound and complete with respect to ⊨,  select all statements that are valid.
We were given a range of options and this was one of the correct answers:


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*W, X, Y ⊨ Z   if and only if   ⊨ W → (X → (Y → Z))

and this was an incorrect answer:

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*X ⊢ Y   if and only if   ⊢ ¬X → ¬Y

I understand what soundness and completeness is
but I do not understand why that option was correct and the second one is incorrect. If someone could explain that would be great
 A: I'm assuming you're working in classical logic (whether propositional logic or first order logic) and using the standard model theoretic semantics and a standard proof system such as tableaux (truth trees), natural deduction or sequent calculus.
First understand the following theorem.
The deduction theorem: If $\Delta\cup \{A\} \vdash B $,  then $\Delta\vdash A\rightarrow B$.
It's also known as "residuation". Informally, it says that if a formula $B$ follows from (is provable from) a set of formulas $\Delta\cup\{A\}$, then $A\rightarrow B$ is provable from $\Delta$ alone. Now what happens if $\Delta$ is the empty set? Then we just have $\{A\}\vdash B$ implies $\vdash A\rightarrow B$.
Secondly, understand that $W,X,Y\vdash Z$ really means $\{W,X,Y\}\vdash Z$. Some authors just leave out the set brackets on the left-hand side of the turnstile symbol and assume that it's understood from context that $W,X,Y$ is viewed as the set that contains $W,X,Y$. However, what appears on the right-hand side of the turnstile symbol is a formula, not a set of formulas.
Why is the first one correct?
I'll cover the "$\Rightarrow$" direction first.
If you have $\{W,X,Y\}\models Z$ then you have $\{W,X,Y\}\vdash Z$ by completeness. Note that  $\{W,X,Y\}=\{W,X\}\cup\{Y\}$ so you have $\{W,X\}\cup\{Y\}\vdash Z$. Now we apply the deduction theorem and get $\{W,X\}\vdash Y\rightarrow Z$. But $\{W,X\}=\{W\}\cup\{X\}$ so we have $\{W\}\cup\{X\}\vdash Y\rightarrow Z$. Applying the deduction theorem gives us $\{W\}\vdash X\rightarrow (Y\rightarrow Z)$. Note that $\{W\}=\emptyset\cup \{W\}$ so we have $\emptyset\cup \{W\}\vdash X\rightarrow (Y\rightarrow Z)$. Apply the deduction theorem again to get $\emptyset\vdash W\rightarrow (X\rightarrow (Y\rightarrow Z))$ which we usually write as $\vdash W\rightarrow (X\rightarrow (Y\rightarrow Z))$. Now, by soundness, we have $\models W\rightarrow (X\rightarrow (Y\rightarrow Z))$.
For the $\Leftarrow$ direction, $\models W\rightarrow (X\rightarrow (Y\rightarrow Z))$ means that $W\rightarrow (X\rightarrow (Y\rightarrow Z))$ is true in every model (every interpretation of the language). In particular, there is no model which makes $W$ true and $X\rightarrow (Y\rightarrow Z)$ false. Thus any model which makes $W$ true, also makes $X\rightarrow (Y\rightarrow Z)$ true. This just means, $W\models X\rightarrow (Y\rightarrow Z)$. Now you can repeat this argument for $X$, then repeat the argument for $Y$ and you end up with $W,X,Y\models Z$.
Why is the second one incorrect?
Either direction can fail. For example, let's consider the $\Rightarrow$ direction. If we have $\vdash X\rightarrow Y$ then by soundness we have $\models X\rightarrow Y$. This just means that $X\rightarrow Y$ is true in every model. In particular, there's no model that makes both $X$ true and $Y$ false. There are models which give some other truth combination for $X$ and $Y$. Consider a model $\mathscr{M}$ which makes both $X$ false and $Y$ true. Note that this model $\mathscr{M}$, doesn't make both $X$ true and $Y$ false. A conditional with a false antecedent is automatically true so $X\rightarrow Y$ is true on $\mathscr{M}$. But notice that  $\neg X\rightarrow \neg Y$ is false on $\mathscr{M}$ because $\mathscr{M}$ makes the antecedent true and the consequent false. In other words, $\mathscr{M}\not\models \neg X\rightarrow \neg Y$. Given that $\neg X\rightarrow \neg Y$ doesn't hold on a particular model $\mathscr{M}$, it cannot hold on every model. So we have $\not\models \neg X\rightarrow \neg Y$. By completeness, we have $\not\vdash \neg X\rightarrow\neg Y$.
