theorem
  (R1+R2).s = R1.s + R2.s
proof
  per cases;
  suppose
A1: not s in Seg i;
    then
A2: not s in dom R2 by FINSEQ_2:124;
A3: not s in dom R1 by A1,FINSEQ_2:124;
    not s in dom(R1+R2) by A1,FINSEQ_2:124;
    hence (R1+R2).s = (0 qua Element of NAT)+(0 qua Element of NAT) by
FUNCT_1:def 2
      .= R1.s + (0 qua Element of NAT) by A3,FUNCT_1:def 2
      .= R1.s + R2.s by A2,FUNCT_1:def 2;
  end;
  suppose
    s in Seg i;
    then s in dom (R1 + R2) by FINSEQ_2:124;
    hence thesis by VALUED_1:def 1;
  end;
end;
