$\mathrm{Halt} = \{ (f,x) | f(x)\downarrow \}$ is r.e. (semi-decidable) but undecidable.$\mathrm{Total} = \{ f | \forall x f(x)\downarrow \}$ is not r.e. (not even semi-decidable).I need some help in proving that $\mathrm{Total}$ is not recursive (decidable). I know the diagonalization proof for the halting problem, I just need the same kind of proof for $\mathrm{Total}$. I'm posting the proof for the halting problem for reference:Undecidability of the halting problemAssume we can decide the halting problem. Then there exists some total function $\mathrm{Halt}$ such that $$\mathrm{Halt}(x,y) = \begin{cases} 1 & \text{if $\phi_x(y)$ is defined}, \\ 0 & \text{if $\phi_x(y)$ is not defined}.\end{cases}$$Here, we have numbered all programs and $\phi_x$ refers to the $x$'th program in this ordering. We can view $\mathrm{Halt}$ as a mapping from $\mathbb{N}$ into $\mathbb{N}$ by treating its input as a single number representing the pairing of two numbers via the one-one onto function $$ \mathrm{pair}(x,y) = \langle x,y \rangle = 2^x (2y + 1)  1 , $$with inverses $$\begin{align*} \langle z \rangle_1 &= \exp(z+1,1), \\ \langle z \rangle_2 &= ((( z + 1 ) // 2^{\langle z \rangle_1})  1 ) // 2 \end{align*}$$Now if $\mathrm{Halt}$ exists, then so does $\mathrm{Disagree}$, where $$\mathrm{Disagree}(x) = \begin{cases} 0 & \text{if $\mathrm{Halt}(x,x)=0$, i.e., if $\phi_x(x)$ is not defined}, \\ \operatorname*{\mu}_y (y=y+1) & \text{if $\mathrm{Halt}(x,x)=1$, i.e., if $\phi_x(x)$ is defined}. \end{cases}$$Since $\mathrm{Disagree}$ is a program from $\mathbb{N}$ into $\mathbb{N}$, $\mathrm{Disagree}$ can be reasoned about by $\mathrm{Halt}$. Let $d$ be such that $\mathrm{Disagree} = \phi_d$, then $$\mathrm{Disagree}(d)\text{ is defined} \Leftrightarrow \mathrm{Halt}(d,d) = 0 \Leftrightarrow _d(d)\text{ is undefined}  \mathrm{Disagree}(d)\text{ is undefined}.$$ But this means that $\mathrm{Disagree}$ contradicts its own existence. Since every step we took was constructive, except for the original assumption, we must presume that the original assumption was in error. Thus, the halting problem is not decidable. 
How to prove that Total is not recursive (decidable)
computability;undecidability;halting problem