What is the subterms of $M\equiv (\lambda x\cdot yx)(\lambda z \cdot x(yx))$? Using this 
Definition (Subterms): The subterms of a term M are defined by induction on $|M|$ as follows:


*

*an atom is a subterm of itself;

*if $M\equiv \lambda x \cdot P$, its subterms are $M$ and all subterms of $P$;

*if $M \equiv P_1 P_2$, its subterms are all the subterms of $P_1$, all those of $P_2$, and $M$ itself.


Example: If $M\equiv (\lambda x\cdot yx)(\lambda z \cdot  x(yx))$ its subterms are $x$, $y$, $yx$, $\lambda x \cdot yx$, $x(yx)$, $\lambda z \cdot x(yx)$ and M itself. (But not $z$.) 
So, my question is: why $z$ is not a subterm in the given example?
 A: It Depends on definition of set of Subterms, If you think $z$ should be a Subterm, you may change the definition as you like:


*

*$\dots$;

*$M \equiv \lambda x \cdot P$, its Subterms are $M$, $x$ and all Subterms of $P$;

*$\dots$.


This may make sense to you as you want $z$ to be a subset of the term you mentioned, but with the definition you provided, it's simple to just go step by step and compute the set of Subterms.
$\text{sub}((\lambda x\cdot yx)(\lambda z\cdot x(yx)))$
(by rule 3.)
$=\{(\lambda x\cdot yx)(\lambda z\cdot x(yx))\} \cup \text{sub}(\lambda x\cdot yx) \cup \text{sub}(\lambda z\cdot x(yx))$
(by rule 2.) (Note here we loose $z$)
$=\{(\lambda x\cdot yx)(\lambda z\cdot x(yx))\} \cup \{\lambda x\cdot yx\} \cup \text{sub}(yx) \cup \{\lambda z\cdot x(yx)\} \cup \text{sub}(x(yx))$
(simplification)
$=\{(\lambda x\cdot yx)(\lambda z\cdot x(yx)), \lambda x\cdot yx, \lambda z\cdot x(yx)\} \cup \text{sub}(yx) \cup \text{sub}(x(yx))$
(by rule 3.)
$=\{(\lambda x\cdot yx)(\lambda z\cdot x(yx)), \lambda x\cdot yx, \lambda z\cdot x(yx)\}  \cup \{yx\} \cup \text{sub}(y) \cup \text{sub}(x) \cup \{x(yx)\} \cup \text{sub}(x) \cup \{yx\} \cup \text{sub}(y) \cup \text{sub}(x)$
(simplification)
$=\{(\lambda x\cdot yx)(\lambda z\cdot x(yx)), \lambda x\cdot yx, \lambda z\cdot x(yx), yx\} \cup \text{sub}(y) \cup \text{sub}(x)  \cup \text{sub}(x) \cup \text{sub}(y) \cup \text{sub}(x)$
(by rule 1.)
$=\{(\lambda x\cdot yx)(\lambda z\cdot x(yx)), \lambda x\cdot yx, \lambda z\cdot x(yx), yx\} \cup \{y\} \cup \{x\} \cup \{x\} \cup \{y\} \cup \{x\}$
(simplification)
$=\{(\lambda x\cdot yx)(\lambda z\cdot x(yx)), \lambda x\cdot yx, \lambda z\cdot x(yx), yx, y, x\}$
