Different definitions of Mathias forcing are equivalent The Mathias Forcing is defined using increasing sequences and I wondered if it would be forcing equivalent if one omits this condition. So, for a free ultrafilter $U$ define
\begin{align*}
   \mathbb{M}_U &= \{ (s,X) \mid X \in U, s \text{ is an increasing finite sequence} \}\\
   {\mathbb{M}_U}^* &= \{ (s,X) \mid X \in U, s \text{ is a finite sequence} \}
\end{align*}
with the usual ordering.
I was able to show that if $f$ is a Mathias real for ${\mathbb{M}_U}^*$, then the enumeration of $\text{range}(f)$ is a Mathias real for $\mathbb{M}_U$. But I got stuck in the other direction, because for example if $f$ is a Mathias real for $\mathbb{M}_U$, then $f$ is necessarily strictly increasing, but any Mathias real $g$ for ${\mathbb{M}_U}^*$ has properties like containing arbitrarily long decreasing sequences, so I do not have an idea how to define such a $g$ from $f$, so I would appreciate a hint how to do the construction if it is even possible.
 A: Thank you Jonathan for the hint, I will elaborate on his comment:
We can show that ${M_U}^*$ always adds a Cohen real. Let $\mathbb{C} = \text{Fn}(\omega, \omega \setminus \{ 0 \})$. Given a Mathias real $f_G$ for ${M_U}^*$ we can decode it to a real $g: \omega \to \omega \setminus \{ 0 \}$ by evaluating the lengths of decreasing sequences in $f_G$. We write $f_G$ codes $g$. We can define $t$ codes $s$ for finite sequences in the same way. Now, define
$$
t ^\frown N \text{ codes } s \iff t \text{ codes } s \text{ and } N > \max(t)
$$
This has the benefit that if $t ^\frown N$ codes $s$ and $f_G \supset t ^\frown N$, then $s \supset g$ as $f_G$ cannot extend the decreasing sequence at the end of $t$ due to $N$ breaking the decreasing sequence.
Now, let $D\in V$ be dense in $\mathbb{C}$ then define
$$
D' = \{ (t ^\frown N, X) \mid X \in U \text{ and } \exists s \in D \text{ such that } t ^\frown N \text{ codes } s \} \in V
$$
We show that $D'$ is dense in ${M_U}^*$, so let $(t,X) \in {M_U}^*$. Then $t$ codes some $s$, so choose $s' \in D$ such that $s' \supseteq s$. But $X$ is infinite, so we can extend $t$ to $t'$ with elements of $X$ such that $t'$ codes $s'$ and choose some $N \in X$ such that $N > \max(t')$. Then $(t' ^\frown N, X) \in D'$ and extends $(t,X)$.
Finally, choose $(t ^\frown N, X) \in G \cap D'$ and choose $s \in D$ such that $t ^\frown N$ codes $s$. Then $f_G \supset t ^\frown N$, so $s \supset g$, which shows that $g$ is a Cohen real.
Now, if $U$ is a selective ultrafilter, then $\mathbb{M}_U$ does not add Cohen reals, so they cannot be forcing equivalent. However, there need not be selective utrafilters, so this does not show that in such models the two definitions are not equivalent.
