theorem Th15:
  x in I & I is proper Ideal of A implies x is NonUnit of A
   proof
     assume that
A1:  x in I and
A2:  I is proper Ideal of A;
     assume
A3:  not x is NonUnit of A;
     reconsider x as Element of A by A1;
     {x}-Ideal = [#]A by A3,RING_2:20; then
     [#]A = I by A1,RING_2:4;
     hence contradiction by A2;
   end;
