The deduction works as follows. For any graph g we transform it into a complete graph G', where the weight of an edge is equal of 1 if this edge already existed in the original set of edges, otherwise the weight of the edge is equal 1 plus cardinality of V time the gap we decided to obtain.
Then we have that if the original graph is a positive instance, that is it admits an Hamiltonian circuit, then clearly the optimal solution for the travelling self-person problem is equal to the cardinality of the set of nodes.
On the other hand, if the graph is a no Hamiltonial circuit then any tour of the travelling self-person has to use at least one node whose cost is 1 + cardinality of V times g. So any tour in the travelling self-person instance must have measure at least equal to cardinality of V-1+1+ cardinality of V times g, which is equal to cardinality V times 1+g.
So we have the gap between the two cases.
