Does $z\mathscr Ex ∧x\mathscr E y ⇒ z\mathscr Ey$ imply $z\mathscr Ex ⇒ z\mathscr Ey$ if $z$ is arbitrary? The proof of Theorem 3 regards $z\mathscr Ex$ $\land$ $x\mathscr E y$ $\Rightarrow$ $z\mathscr Ey$ implies $z\mathscr Ex \Rightarrow z\mathscr E y$. But I think it's not correct because the hypothesis $z\mathscr Ex$ $\land$ $x\mathscr E y$ and $z\mathscr Ex$ is different.
"Proof of Theorem 3 

(c) It follows immediately from (a) and (b) above that x/$\mathscr E$ = y/$\mathscr E \Rightarrow x \mathscr E$ y
We need to prove that x$\mathscr E$ $y\Rightarrow x/\mathscr E = y/\mathscr E$ 
Let x$\mathscr E$y. Then
  $z\in x/\mathscr E \Rightarrow z\mathscr E x$   Def. 6        

$\space\space\color{gray} {\text{The author repeats the conjunction of above two hypothesises in symbols as the following}}$  

$\underline {z\mathscr Ex \land x\mathscr E y \Rightarrow z\mathscr Ey}$ $\space\space\space\space\space\mathscr E$ is transitive.
  $\underline {\Rightarrow z\in y/\mathscr E}$ $\space\space\space\space\space Def. 6$   

$\color{gray}{\text {Now the author regards the above conclusion $z\mathscr E x \land x\mathscr E y \Rightarrow z\mathscr E y$ is equivaelnt to $z\mathscr E x \Rightarrow z\mathscr E y$ without revealing the justification to the reasoning}}$
$\color{green}{\text {Now I understand $z\mathscr E x \land x\mathscr E y \Rightarrow z\mathscr E y\equiv z\mathscr E x \Rightarrow (x\mathscr Ey\Rightarrow z\mathscr E y)$ by Exportation law in Example 7}}$ 

Since $z$ is arbitrary, it follows that $x/\mathscr E \subseteq y/\mathscr E$. A similar argument gives $y/\mathscr E \subseteq x/\mathscr E$; hence $x/\mathscr E = y/\mathscr E$
  "
  Q.E.D.

FYI

"Theorem 3. Let $\mathscr E$ be an equivalence relation on a nonempty set $X$. Then
  (a) Each $x/\mathscr E$ is a nonempty subset of $X$.
  (b)  $x/\mathscr E \cap y/\mathscr E \neq \emptyset$ if and only if $x\mathscr Ey$.
  (c) $x\mathscr Ey$ if and only if $x/\mathscr E = y/\mathscr E$"
  ...
Definition 6. Let $\mathscr E$ be an equivalence relation on a nonempty set $X$. For each $x\in X$, we define
​$X/\mathscr E=\{\,y∈X\mid y\mathscr E x\,\}$
which is called the equivalence class determined by the element $x$.
The set of all such equivalence classes on $X$ is denoted by $X/\mathscr E$; that is, $X/\mathscr E=\{\,x/\mathscr E\mid x∈X\,\}$. 
The symbol $X/\mathscr E$ is read "$X$ modulo $\mathscr E$," or simply "$X$ mod $\mathscr E$".
Example 7 Prove the following Exportation Law:
  $\space\space\space\space\space p \land q \to r \equiv p \to (q\to r)$
[Solution] $p\to (q \to r) \equiv p \to$ ~($q\land$ ~r)    Def. 4
  $\space\space\space\space\equiv$ ~[$p\land(q\land$ ~r)]$\space\space Def. 4, D.N.$
  $\space\space\space\space\equiv$ ~[($p\land q)\land$~r]$\space\space$ Assoc.
  $\space\space\space\space\equiv p\land q\to r\space\space$ Def. 4  
Hence, p$\land q \to r \equiv p \to (q \to r)$
Definition 4
  The connective $\rightarrow$ is called the conditional and may be placed between any two statement p and q to form the compound statement p→q(reaad: "if p, then q". By definition the statement p→q is equivalent to the statement ~(p∧~q).

Source: Set Theory by You-Feng Lin, Shwu-Yeng T. Lin
 A: We need the definition of valid argument [page 24-25]:

A formal proof of validity for a given argument is a sequence of statements each of which is either a premise of the argument or follows from preceeding statments by a known valid argument, ending with the conclusion of the argument. 

The proof begins with :
H1. $x\mathscr Ey$ --- premise of the argument
H2. $z\in x/\mathscr E$ --- 2nd premise of the argument
Def.6 : $z\in x/\mathscr E \Rightarrow z\mathscr E x$
Thus, by Modus Ponens [page 20], we have:
C. $z\mathscr E x$.
Up to now, we have proved:

H1. $x\mathscr Ey$ and H2. $z\in x/\mathscr E$, therefore C. $z\mathscr E x$.

We can symbolize it as : $P,Q \Rightarrow R$. 
Clearly: $P \Rightarrow P$ [see page 12 : $P \to P$ is a tautology] and thus, by properties of implies: $P,Q \Rightarrow P$.
By Conjunction [page 25]: $R,P \Rightarrow R \land P$.
Thus, putting all together, we have : $P, Q \Rightarrow R \land P$, i.e.:

$x\mathscr Ey$ and $z\in x/\mathscr E$, therefore $z\mathscr E x \land x\mathscr Ey$.

By transitivity of $\mathscr E$: $z\mathscr Ex \land x\mathscr E y \Rightarrow z\mathscr Ey$ and thus:

$x\mathscr Ey$ and $z\in x/\mathscr E$, therefore $z\mathscr Ey$.

Finally, we apply again Def.6 to $z\mathscr Ey$ in order to derive:

$x\mathscr Ey$ and $z\in x/\mathscr E$, therefore $z\in y/\mathscr E$.

Thus, applying Exportation [see page 19: $[(P \land Q) \to R] \equiv [P \to (Q \to R)]$ ] we can conclude with :


$x\mathscr Ey$, therefore: if $z\in x/\mathscr E$, then $z\in y/\mathscr E$,


i.e.

$x\mathscr Ey \Rightarrow [(z\in x/\mathscr E) \to (z\in y/\mathscr E)]$.

Now we need the so-called Universal Generalization rule ["since $z$ is arbitrary, it follows that ..."; it seems to me that it is not discussed in the book], to conclude with:


$x\mathscr Ey \Rightarrow (\forall z) [(z\in x/\mathscr E) \to (z\in y/\mathscr E)]$.


Now, by Def.2 [page 36], we have:


$x\mathscr Ey \Rightarrow x/\mathscr E \subseteq y/\mathscr E$.


A: At that point in the text, you already know (i.e., have assumed that) $x {\mathscr E} y$. Under that assumption, $z {\mathscr E} x$ indeed implies (and is equivalent to) $z {\mathscr E} y$.
A: For part c), two things need to be proven:


*

*$ x/\mathscr E=y/\mathscr E\implies x\mathscr E y$

*$x\mathscr E y\implies x/\mathscr E=y/\mathscr E$


The first implication is covered by the  "It follows immediately ..."
To prove the second direction, the author starts with the assumption that $ x\mathscr E y$ and the goals is to prove $ x/\mathscr E=y/\mathscr E$. To show equality of sets one can proceed by showing both  $x/\mathscr E\subseteq y/\mathscr E$ and  $y/\mathscr E\subseteq x/\mathscr E$.
To show $x/\mathscr E\subseteq y/\mathscr E$, one starts with an arbitrary element $z\in x/\mathscr E$ and intends to show that  $z\in y/\mathscr E$.
Now, $z\in x/\mathscr E$ means precisely that $z\mathscr E x$. This, together with the initial assumption $x\mathscr E y$ entails $z\mathscr E y$ (as $\mathscr E$ is transitive).
That again means $z\in y/\mathscr E$ as desired.
To summarize the last step: We have shown $\forall z\colon z\in x/\mathscr E\to  z\in y/\mathscr E$, which is just a rewording of $x/\mathscr E\subseteq y/\mathscr E$.
