theorem
  z in dom f & f.z = x implies (f+~(x,y)).z = y
proof
  assume that
A1: z in dom f and
A2: f.z = x;
  f.z in dom(x.-->y) by A2,FUNCOP_1:74;
  then
A3: z in dom((x.-->y)*f) by A1,FUNCT_1:11;
  hence (f+~(x,y)).z = ((x.-->y)*f).z by Th13
    .= (x.-->y).x by A2,A3,FUNCT_1:12
    .= y by FUNCOP_1:72;
end;
