theorem Th27:
  (X-->0c) + (X-->0c) = X-->0c & a(#)(X-->0c) = X-->0c
proof
  set g1 = X-->0c;
  set g2 = X-->0c;
A1: now
    let x be Element of X;
    assume x in dom(a(#)g1);
    then (a(#)g1).x = a * g1.x by VALUED_1:def 5;
    then (a(#)g1).x = a * 0 by FUNCOP_1:7;
    hence (a(#)g1).x = (X --> 0c).x by FUNCOP_1:7;
  end;
A2: dom(g1+g2) = dom g1 /\ dom g2 by VALUED_1:def 1;
  now
    let x be Element of X;
    assume x in dom(X --> 0c);
    then (g1+g2).x = g1.x + g2.x by A2,VALUED_1:def 1;
    then (g1+g2).x = 0 + g2.x by FUNCOP_1:7;
    hence (g1+g2).x = (X --> 0c).x;
  end;
  hence g1+g2 = X-->0c by A2,PARTFUN1:5;
  dom(a(#)g1) = dom(X-->0c) by VALUED_1:def 5;
  hence thesis by A1,PARTFUN1:5;
end;
