Proving a certain superset the halting language is not recursive
Main idea of the proof is since $L$ is kind of superset of $L_{halt}$ we select reduction in such a way that we always deal with only that subset of $L$ which resembles $L_{halt}$.

First let us choose any encoding such that if $\langle M,w \rangle \in L_{halt}$ then $\langle M,w \rangle^R \not\in L_{halt}$. $L_{halt}$ is undecidable in every reasonable encoding so this is okay. Also this will be true for most of the encodings if we include $\langle \rangle$ also in the encoding. (Reverse will be $\rangle w^R, M^R \langle$). Also we should have $val(\langle) &gt; val(\rangle)$. I assume that $val$ is calculated as a number with base $|\Gamma|$ where $\Gamma$ is the tape alphabet. In the question decimal base is mentioned, which means the same. We need to add extra 4 bits (= 1 decimal digit) in each cell if we do not have enough space for new symbols.

For this encoding we can have an identity reduction for valid input of the form $\langle M,w \rangle$. For invalid input not of the form $\langle M,w \rangle$ we shall reduce the input to $\langle\epsilon,\epsilon\rangle$ which does not belong to $L$. Let us call this reduction $f$. We assume that in our encoding $\langle\epsilon,\epsilon\rangle \not\in L_{halt}$.

We can now easily prove $x \in L_{halt}$ if and only if $f(x) \in L$.

($\Rightarrow$)
Let $x = \langle M,w \rangle \in L_{halt}$.
Then clearly $val(x) &gt; val(x ^R)$.
And $\langle M,w \rangle^R \not\in L_{halt}$.
Therefore $\langle M,w \rangle \in L_3$ and hence $\langle M,w \rangle \in L$ as naturally $x \neq x^R$, Symbols $\langle,\rangle$ ensures that $x \neq x^R$.

($\Leftarrow$)   

Note it $f(x) = \langle \epsilon, \epsilon \rangle$ then it does not belong to $L$. So take the other valid cases.

Let $f(x)= y =\langle M,w \rangle \in L$.
Then $\langle M,w \rangle \in L_3$, as $val(y) &gt; val(y^R)$.
This gives us $\langle M,w \rangle \in L_{halt}$ straight away.

The idea of this proof holds even if we are given a specific encoding for $L$ as well.

1) If the encoding for $L$ is such that we do not have begin symbol $\langle$ and end symbol $\rangle$. Also assume $x_{reject}$ is a string such that $x_{reject} \not\in L$. We can easily find such a string as $L \neq \Sigma^*$. In this case the strings got by reduction will not be necessarily in $L_3$. We also assume our encoding of turing machines is such that we can find a string $\alpha$ such that $0^*\alpha$ is never a proper prefix.

To always get strings in $L_3$ we define the reduction as follows. All invalid input strings which are not of the form $\langle M,w \rangle$ are reduced to $x_{reject}$, and all valid input strings which are of the form $\langle M,w \rangle$ are reduced to $\langle M',w\alpha^R0^{|M'|+1}\rangle$. $M'$ will remove suffix $\alpha^R0^*$ and then behave as $M$. 

With this reduction the proof given above can be modified slightly to prove that $L$ is undecidable.

If the encoding for $L$ is such that $\langle$ and $\rangle$ are always included and $val(\langle) &lt; val(\rangle)$ then in the reductions given above we will always get strings from set $L_1$ and $L_2$. In this case by using $\alpha^R0^*$ suffix it is easier to get reduced strings from set $L_2$. Then it is better to reduce language $\overline{L_{halt}}$ which is also undecidable to $L$ to prove undecidability of $L$.