
theorem Th13:
  for n be Element of NAT holds Decomp(n,2).1 = <*0,n*>
proof
  let n be Element of NAT;
  consider A be finite Subset of 2-tuples_on NAT such that
A1: Decomp(n,2) = SgmX (TuplesOrder 2,A) and
A2: for p be Element of 2-tuples_on NAT holds p in A iff Sum p = n by Def4;
A3: now
    let y be Element of 2-tuples_on NAT;
    consider d1,d2 be Element of NAT such that
A4: y = <*d1,d2*> by FINSEQ_2:100;
    assume y in A;
    then Sum <*d1,d2*> = n by A2,A4;
    then
A5: d1 + d2 = n by RVSUM_1:77;
    now
      per cases;
      suppose
        d1 = 0;
        hence <*0,n*> <= <*d1,d2*> by A5;
      end;
      suppose
        d1 > 0;
        then <*0,n*>.1 < d1;
        then
A6:     <*0,n*>.1 < <*d1,d2*>.1;
        1 in Seg 2 & for k be Nat st 1 <= k & k < 1 holds <*0,
        n*>.k = <* d1,d2*>.k by FINSEQ_1:1;
        then <*0,n*> < <*d1,d2*> by A6;
        hence <*0,n*> <= <*d1,d2*>;
      end;
    end;
    hence [<*0,n*>,y] in TuplesOrder 2 by A4,Def3;
  end;
  1 <= n+1 by NAT_1:11;
  then 1 in Seg (n+1) by FINSEQ_1:1;
  then 1 in Seg len Decomp(n,2) by Th9;
  then
A7: 1 in dom Decomp(n,2) by FINSEQ_1:def 3;
  field TuplesOrder 2 = 2-tuples_on NAT by ORDERS_1:15;
  then
A8: TuplesOrder 2 linearly_orders A by ORDERS_1:37,38;
  Sum <*0,n*> = 0 qua Nat+n by RVSUM_1:77;
  then <*0,n*> in A by A2;
  then (SgmX (TuplesOrder 2,A))/.1 = <*0,n*> by A8,A3,PRE_POLY:20;
  hence thesis by A1,A7,PARTFUN1:def 6;
end;
