
theorem Th8:
  for n be Element of NAT holds len Decomp(n,1) = 1
proof
  let n be Element of NAT;
  consider A be finite Subset of 1-tuples_on NAT such that
A1: Decomp(n,1) = SgmX (TuplesOrder 1,A) and
A2: for p be Element of 1-tuples_on NAT holds p in A iff Sum p = n by Def4;
A3: A = {<*n*>}
  proof
    thus A c= {<*n*>}
    proof
      let y be object;
      assume
A4:   y in A;
      then reconsider p=y as Element of 1-tuples_on NAT;
A5:   Sum p = n by A2,A4;
      consider k be Element of NAT such that
A6:   p = <*k*> by FINSEQ_2:97;
      Sum p = k by A6,RVSUM_1:73;
      hence thesis by A5,A6,TARSKI:def 1;
    end;
    let y be object;
    assume y in {<*n*>};
    then
A7: y = <*n*> by TARSKI:def 1;
    Sum <*n*> = n by RVSUM_1:73;
    hence y in A by A2,A7;
  end;
  field TuplesOrder 1 = 1-tuples_on NAT by ORDERS_1:15;
  then TuplesOrder 1 linearly_orders 1-tuples_on NAT by ORDERS_1:37;
  hence len Decomp(n,1) = card A by A1,ORDERS_1:38,PRE_POLY:11
    .= 1 by A3,CARD_1:30;
end;
