Using diagonal argument to prove that $H(x) = \mu y T(x,x,y)$ has no total computable extension Hello everyone just like the title says I want to prove that $H(x) = \mu y T(x,x,y)$ has no total computable extension such that if we had a function $BIG(x)$ that is both total and agrees with $H(x)$ whenever $H(x)$ is defined, then $BIG(x)$ is not computable. This is a homework question so I don't want a full solution just some help! 

$\bf{NOTE}:$ The predicate $T(y,x,z)$ means that it holds iff program $y$ takes an input $x$ (could be $n$-ary) and has a computational history z! This is supposed to be the Kleene T predicate basically.  

The function $H(x)$ I believe returns the smallest computational history $y$ such that a program $\{x\}(x)$ (program takes input of its own configuration and runs) runs and halts, since $\mu y R(x,y)$ means the smallest $y$ such that $R(x,y)$ holds. Maybe I am not quite clear what it means for $BIG$ to agree with $H(x)$ or what it's own input. I think I need to create a diagonal function that uses $BIG$ if $BIG$ was computable and show that if I had some program $e$ then it must agree with $BIG$ but based on my definition of that diagonal function it isn't. If you are reading this you might see the mess of my thinking, any help would be greatly appreciated!
(This answer assumes my conjectural interpretation of the predicate $T$.)
The function $H(x)$ is given by
$$
H(x) = \begin{cases} t &amp; \text{if program $x$ run on input $x$ halts after $t$ steps} \\ \bot &amp; \text{if program $x$ never halts when run on input $x$} \end{cases}
$$

Hint: Given a total computable function extending $H$, show how to solve the halting problem. There is no need to use diagonalization – this is already taken care of in the proof that the halting problem is undecidable.