theorem Th15:
  for x,y holds (x '=' y).1 = 0 & (x 'in' y ).1 = 1
proof
  let x,y;
  thus (x '=' y).1 = (<*0*>^(<*x*>^<*y*>)).1 by FINSEQ_1:32
    .= 0 by FINSEQ_1:41;
  thus (x 'in' y).1 = (<*1*>^(<*x*>^<*y*>)).1 by FINSEQ_1:32
    .= 1 by FINSEQ_1:41;
end;
