Let TOT be the set of all Turing Machines that halt on all inputs. Find a computable set B of ordered triples such that:

TOT = {e : ($$\forall$$x)($$\exists$$y)[(e, x, y) $$\in$$ B]

This definition means that TOT is a set of all Turing machines e such that they halt on all inputs. The "for all" x denotes all inputs to that machine, and "there exists" a y denotes that e halts under y steps. x consists of 0s and 1s, y and e are Natural numbers too ( e denotes Turing machine $$T_e$$ if we were to number all our turing machines)

EDIT: SOLVED

• What do $e$, $x$, and $y$ represent? I guess that $e$ is a Turing machine, to be coherent with the definition of TOT. What is $x$ and what is $y$? May 10 '20 at 15:32
• The definition of TOT is clear to me. From your comment, I get that $x$ belongs to $\{0,1\}^*$. It is still not clear the set to which $y$ belongs to. Is it an integer? A word in \{0, 1\}^*? The condition "Turing machine e halts under y steps" does not appear at all in your problem statement... May 10 '20 at 15:45

Let $$T$$ be the set of all Turing machines.
Let $$B = \{ (e,x, y) \in T \times \{0,1\}^* \times \mathbb{N} : e(x) \text{ halts in y steps}\}$$, and define $$C = \{e \in T : \forall x \in \{0,1\}^*, \; \exists y \in \mathbb{N}, \; (e, x, y) \in B]$$.
We start by proving $$\text{TOT}=C$$.
If $$e \in \text{TOT}$$ then, $$\forall x \in \{0,1\}^*$$, $$e(x)$$ halts. Let $$y_x$$ be the number of steps needed for $$e(x)$$ to halt. Then $$\forall x \in \{0,1\}^*$$ there is a choice of $$y \in \mathbb{N}$$, namely $$y = y_x$$, such that $$(e,x,y) \in B$$, showing that $$e \in C$$ and hence $$\text{TOT} \subseteq C$$.
If $$e \not\in \text{TOT}$$, then $$\exists x \in \{0,1\}^*$$ such that $$e(x)$$ does not halt. Then $$\exists x$$, such that $$\forall y \in \mathbb{N}$$, $$(e, x, y) \not\in B$$, showing that if is false that $$\forall x \in \{0,1\}^*, \exists y \in \mathbb{N}$$ such that $$(e, x, y) \in B$$. As a consequence $$e \not\in C$$ implying $$\text{TOT} \supseteq C$$.
It remains to show that $$B$$ is computable. To decide whether any given triple $$(e, x,y) \in T \times \{0,1\}^* \times \mathbb{N}$$ belongs to $$B$$ it suffices to consider a Turing machine $$T'$$ that simulates $$e(x)$$ for $$y$$ steps. If $$e(x)$$ halts, then $$T'$$ accepts. Otherwise, if $$e(x)$$ does not halt by the end of the $$y$$-th step, $$T'$$ rejects.