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In theory of computation, if a language $L_{1_{}}$is decided by a Turing Machine based on another language $L_{2_{}}$, then we say that A can be decidable to $L_{2_{}}$, or $L_{1_{}}$is Turing Reducible to $L_{2_{}}$. It is denoted as $L_1{鈮�}_T{L}_2$.

In this question, we need to prove that, 鈥淎 is Turing Reducible to B but B but is not Turing Reducible to A鈥�.

Let $A=E_{T{M}_{}}=Emty$Turing Machine ={<M>| M is a Turing Machine and language L$\left(M\right)=蠒\}$

And B = $A_{T{M}_{}}$= Oracle Turing Machine Machine ={<,w>| Machine M accept string w}

Let us first prove $A{鈮�}_{T}B$

To prove it, we need to prove $E_{TM}{鈮�}_T{A}_{TM}$

We know that $E_{TM}$is decidable relative to $A_{TM}$.

Since, above is true, that means A is decidable relative to B then A ${鈮�}_T$B.

Now we will prove that B' ${}_{T}A$

To prove this, we need to prove $A_{TM}{\text{'}}_T{E}_{TM.}$.

Let us assume that $A_{TM}\text{'}{}_T{E}_{TM.}$.

Now by mapping reducibility rules for any two language : language $x鈮�my<x{鈮�}_{m}y:Languagex$is map reducible to language y .

If there is a computable function f : $鈭�*鈫�鈭�*$

Here, for every w, function f called reduction of x to y .

This means, w e x, f(w) e y

From above reduce mapping, we can conclude that $A_{TM}{鈮�}_T{E}_{TM},A_{TM}{鈮�}_M{E}_{TM}$

But, $E_{TM}$is Turning Recognizable but $A_{TM}$ is not.

This violates mapping reducibility as both must be Turing Recognizable.

So, our assumption was wrong that .$A_{TM}{鈮�}_T{E}_{TM}$

Hence, B'${}_{T}A$

Therefore, we conclude that: and $A{鈮�}_{T}B$and $B\text{'}{}_{T}A.$'.