If $M$ is a transitive set and $R$ well orders $A$ in $M$, then $R$ well orders $A$ universally

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I am asked to prove the following: Working in ZF - P.

If $M$ is (Edit) a transitive model of ZF - P, and $R, A \in M$, then

$(R \ well \ orders \ A) ^M \rightarrow (R \ well \ orders \ A)$

I am struggling with the minimal element. Since $R$ well orders $$A relative to $M$ we have that: $\forall X \in M s.t. X \subset A (X \neq 0 \rightarrow \exists y \in X (\lnot \exists z \in X (zRy))]$

So I need to show that for any $X$ (not necessarily in $M$) s.t. $X \subset A$ and $X \neq 0$, X has a minimal element.

My idea was to suppose there is such an X with no minimal element and arrive at a contradiction by showing this would imply A has no minimal element but the details of this are tripping me up.

Considering any element $a \in A$ we try to show $\exists b \in A$ s.t. $b R a$.

If $a \in A$ then $a \in X$ or $a \in A-X$. If $a \in X$ then we are done since X has no minimal element so we can always find something smaller.

Now if $ a \in A-X$ then...I want to say something in $X$ is smaller then it? Or possibly compare $a$ and elements in $X$ since those are both in $M$ and hence well ordered there, but I'm getting stuck.

Any assistance would be awesome.