reserve x,y for object, X for set;

theorem Th9:
  for n,m,k be non zero Nat st k = n lcm m holds
  support ppf k=(support ppf n) \/ (support ppf m)
proof
  let n,m,k be non zero Nat;
  assume
A1: k = n lcm m;
A2: support ppf n = support pfexp n by NAT_3:def 9;
A3: support pfexp n \/ support pfexp m c= support max(pfexp n,pfexp m)
  proof
    let x be object;
    assume
A4: x in support pfexp n \/ support pfexp m;
    per cases by A4,XBOOLE_0:def 3;
    suppose
A5:   x in support pfexp n;
A6:   (pfexp n).x <=max(pfexp n,pfexp m).x
      proof
        per cases;
        suppose
          (pfexp n).x <= (pfexp m).x;
          hence thesis by NAT_3:def 4;
        end;
        suppose
          (pfexp n).x > (pfexp m).x;
          hence thesis by NAT_3:def 4;
        end;
      end;
      (pfexp n).x <> 0 by A5,PRE_POLY:def 7;
      then 0 < (pfexp n).x;
      hence thesis by A6,PRE_POLY:def 7;
    end;
    suppose
A7:   x in support pfexp m;
A8:   (pfexp m).x <=max(pfexp n,pfexp m).x
      proof
        per cases;
        suppose
          (pfexp n).x <= (pfexp m).x;
          hence thesis by NAT_3:def 4;
        end;
        suppose
          (pfexp n).x > (pfexp m).x;
          hence thesis by NAT_3:def 4;
        end;
      end;
      (pfexp m).x <> 0 by A7,PRE_POLY:def 7;
      then 0 < (pfexp m).x;
      hence thesis by A8,PRE_POLY:def 7;
    end;
  end;
A9: support ppf m = support pfexp m by NAT_3:def 9;
A10: support ppf k =support pfexp k by NAT_3:def 9
    .=support max(pfexp n,pfexp m) by A1,NAT_3:54;
  then support ppf k c= (support ppf n) \/ (support ppf m) by A2,A9,NAT_3:18;
  hence thesis by A10,A2,A9,A3,XBOOLE_0:def 10;
end;
