theorem Th31:
  for x being Element of [: base_of_frechet_filter,base_of_frechet_filter :]
  holds ex i,j st x = [:NAT \ Segm i,NAT \ Segm j:]
  proof
    let x be Element of [: base_of_frechet_filter,base_of_frechet_filter :];
    x in the set of all [:B1,B2:] where B1 is Element of
    base_of_frechet_filter, B2 is Element of base_of_frechet_filter;
    then consider B1 be Element of base_of_frechet_filter,
                  B2 be Element of base_of_frechet_filter such that
A1: x = [:B1,B2:];
    consider i such that
A2: B1 = NAT \ Segm i by Th20;
    consider j such that
A3: B2 = NAT \ Segm j by Th20;
    take i,j;
    thus thesis by A1,A2,A3;
  end;
