theorem Th3:
  { x1,x2,x3 } = { x1,x2 } \/ { x3 }
proof
  thus { x1,x2,x3 } = { x1 } \/ { x2,x3 } by Th2
    .= { x1 } \/ ({ x2 } \/ { x3 }) by Th1
    .= { x1 } \/ { x2 } \/ { x3 } by XBOOLE_1:4
    .= { x1,x2 } \/ { x3 } by Th1;
end;
