theorem Th16:
  B0 is R-normal & |.x0.| = 1 implies B0 \/ {x0} is R-normal
proof
  assume that
A1: B0 is R-normal and
A2: |.x0.| =1;
  {x0} is R-normal by A2,Th15;
  hence B0 \/ {x0} is R-normal by A1;
end;
