Can anyone make me understand in simple language why the second condition for being an Euclidean domain is superfluous? Can anyone make me understand in simple language why the second condition for being an Euclidean domain is superfluous ?

Why $v(a)  \leq v(ab)$ is not needed?  How we can deduce from the first one?
Suppose $V$ satisfies only the first Euclidean property, i.e. for all $\,a,b\in D\,$ if $\,b\neq 0\,$ then there are $\,q,r\in D\,$ such that $\, a = qb +r\,$ with $\, V(r) &lt; V(b),\,$ where $V$ maps $D$ into  (well-ordered)  $\,\Bbb N.\,$ We show how to construct from $V$ another Euclidean function $v$ that satisfies $\, v(a) \le v(ab)\,$ if $\,ab\neq 0$.
Derive $\,v\,$ from $\,V\,$ as follows
$$\begin{align} 
v(0) &amp;= V(0)\\
v(a) &amp;= {\rm min}\{ V(b)\ :\ b\in aD\backslash 0\}
\end{align}$$
Note $\,v(a)\le V(a)\,$ since it is clear if $\,a = 0,\,$ else  it follows by $\, a\in aD\backslash 0$
$v$ is also a Euclidean function: if $\,a,b\in D\,$ and $\,b\neq 0\,$ then $\,v(b) = V(bc)\,$ for $\,0\neq c\in D.\,$  Since $\,V\,$ is a Euclidean function there are $\,q,r\in D\,$ such that $\, a = qbc + r\,$ and $\,V(r) &lt; V(bc) = v(b).\,$ But by above we know $\,v(r)\le V(r)\,$ thus $\,v(r) &lt; v(b),\,$ so $\,v\,$ is a Euclidean function.
Note $\, v(a) \le v(ab)\,$ if $\,ab\neq 0\,$ since $\,aD\backslash 0\supseteq abD\backslash 0$ $\,\Rightarrow\,{\rm min}\,V(aD\backslash 0) \le {\rm min}\,  V(abD\backslash 0)$
Remark $ $ See the paper cited here by Agargun &amp; Fletcher for a comprehensive study of the logical relationships between various common definitions of Euclidean domains and rings.
The second property is superfluous because only the first one is needed to prove that every ideal of a Euclidean domain is principal.