theorem Th45:
  v in V & a in A implies ND_ex_1(v,a) in NDSS(V,A)
  proof
    assume that
A1: v in V and
A2: a in A;
    reconsider V1 = V, A1 = A as non empty set by A1,A2;
    reconsider v as Element of V1 by A1;
    reconsider a as Element of A1 by A2;
    v.-->a in NDSS(V,A);
    hence thesis;
  end;
