reserve n for Element of NAT,
  i for Integer,
  a, b, r for Real,
  x for Point of TOP-REAL n;

theorem
  b-a <= 1 implies IntIntervals(a,b) is mutually-disjoint
proof
  assume
A1: b-a <= 1;
A2: now
    let k be Element of NAT;
A3: a+1+0 <= a+1+k by XREAL_1:6;
    b-a+a <= 1+a by A1,XREAL_1:6;
    hence a+(k+1) >= b by A3,XXREAL_0:2;
  end;
  let x, y be set;
  assume x in IntIntervals(a,b);
  then consider nx being Element of INT such that
A4: x = ].a+nx,b+nx.[;
  assume y in IntIntervals(a,b);
  then consider ny being Element of INT such that
A5: y = ].a+ny,b+ny.[;
  assume
A6: x <> y;
  assume x meets y;
  then consider z being object such that
A7: z in x and
A8: z in y by XBOOLE_0:3;
  reconsider z as Real by A4,A7;
A9: a+nx < z by A4,A7,XXREAL_1:4;
A10: z < b+ny by A5,A8,XXREAL_1:4;
A11: a+ny < z by A5,A8,XXREAL_1:4;
A12: z < b+nx by A4,A7,XXREAL_1:4;
  per cases by XXREAL_0:1;
  suppose
    nx = ny;
    hence contradiction by A4,A5,A6;
  end;
  suppose
    nx < ny;
    then nx+1 <= ny by INT_1:7;
    then reconsider k = ny-(nx+1) as Element of NAT by INT_1:5;
    a+nx+1+k < b+nx by A12,A11,XXREAL_0:2;
    then a+nx+(k+1)-nx < b+nx-nx by XREAL_1:14;
    then a+(k+1) < b;
    hence contradiction by A2;
  end;
  suppose
    nx > ny;
    then ny+1 <= nx by INT_1:7;
    then reconsider k = nx-(ny+1) as Element of NAT by INT_1:5;
    a+ny+1+k < b+ny by A9,A10,XXREAL_0:2;
    then a+ny+(k+1)-ny < b+ny-ny by XREAL_1:14;
    then a+(k+1) < b;
    hence contradiction by A2;
  end;
end;
