theorem Th31:
  F is additive & (for n holds G.n = (F.n)|D) implies G is additive
proof
  assume that
A1: F is additive and
A2: for n holds G.n = (F.n)|D;
  let n,m be Nat;
A3: G.m = (F.m)|D by A2;
  then
A4: dom(G.m) c= dom(F.m) by RELAT_1:60;
  assume n <> m;
  let x be set;
  assume
A5: x in dom(G.n) /\ dom(G.m);
  then
A6: x in dom(G.m) by XBOOLE_0:def 4;
A7: G.n = (F.n)|D by A2;
  then dom(G.n) c= dom(F.n) by RELAT_1:60;
  then dom(G.n) /\ dom(G.m) c= dom(F.n) /\ dom(F.m) by A4,XBOOLE_1:27;
  then
A8: (F.n).x <> +infty or (F.m).x <> -infty by A1,A5;
  x in dom(G.n) by A5,XBOOLE_0:def 4;
  hence thesis by A7,A3,A8,A6,FUNCT_1:47;
end;
