reserve a,b,p,x,x9,x1,x19,x2,y,y9,y1,y19,y2,z,z9,z1,z2 for object,
   X,X9,Y,Y9,Z,Z9 for set;
reserve A,D,D9 for non empty set;
reserve f,g,h for Function;
reserve A,B for set;
reserve x,y,i,j,k for object;

theorem
  for f be Function, a,b,n,m,x be object st x <> m & x <> a
   holds (f +* (a.--> b) +* (m .--> n)).x=f.x
proof
  let f be Function, a,b,n,m,x be object;
  assume that
A1: x<>m and
A2: x<>a;
  set mn=m .--> n, nm=a .--> b;
A3: not x in dom nm by A2,TARSKI:def 1;
  not x in dom mn by A1,TARSKI:def 1;
  hence (f +* nm +* mn).x=(f +* nm).x by Th11
    .=f.x by A3,Th11;
end;
