Let $PAL = \{ww^R\ | w\in\{0,1\}^*\}$.
Then let $A = \{\langle M\rangle \ | \textit{M is a Turing Machine and } L(M)\subseteq PAL\}$
Is A semi-decidable (Turing recognizable or recursively enumerable)?
My try!
I said it is so here is my proof, tell me if I am way off base or tell me if A ought not to semi-decidable in the first place.
$\underline{Proof}$
A is semi-decidable by The Certificate Theorem, which states that:
If $A\subseteq\Sigma^*$ then $A\in SD$(semi-decidable) iff there is some decidable relation, $R\subseteq \Sigma^* \times \Sigma^*$, such that $\forall x\in\Sigma^*$, $x\in{A}$ iff $\exists y$ such that $R(x,y)$ holds.
We know that if $x\in{A}$ that means that $x=\langle M\rangle$, so we can choose $y = \langle M,w\rangle$ where $w\in{PAL}$. What we know is that when $M$ takes $w$ as an input it will accept (since we choose $w$ as an input), if that is the case then $w\in{L(M)}\subseteq{PAL}$. That means that $A\in{SD}$. QED
Any help would be greatly appreciated, have a great day!