theorem Th56:
  f c= g implies sproduct f c= sproduct g
proof
  assume
A1: f c= g;
  then
A2: dom f c= dom g by GRFUNC_1:2;
  let y be object;
  assume y in sproduct f;
  then consider h such that
A3: y = h and
A4: dom h c= dom f and
A5: for x being object st x in dom h holds h.x in f.x by Def9;
A6: dom h c= dom g by A2,A4;
  now
    let x be object;
    assume
A7: x in dom h;
    then f.x = g.x by A1,A4,GRFUNC_1:2;
    hence h.x in g.x by A5,A7;
  end;
  hence thesis by A3,A6,Def9;
end;
