reserve n,i,j,k for Nat;

theorem Th4:
  for x1,x2 being Element of NAT holds Sum Prefix(<*x1,x2*>,1)=x1 &
  Sum Prefix(<*x1,x2*>,2)=x1+x2
proof
  let x1,x2 be Element of NAT;
  reconsider y1=x1 as Element of INT by INT_1:def 2;
  thus Sum Prefix(<*x1,x2*>,1)=Sum<*y1*> by FINSEQ_6:3
    .=x1 by FINSOP_1:11;
  reconsider y2=x2 as Element of INT by INT_1:def 2;
  len <*x1,x2*>= 2 by FINSEQ_1:44;
  hence Sum Prefix(<*x1,x2*>,2)=Sum<*y1,y2*> by FINSEQ_3:49
    .=x1+x2 by RVSUM_1:77;
end;
