reserve A for QC-alphabet;

theorem Th3:
  for x,y being set, g being Function, A being set holds (g +* (x
  .--> y)).:(A \ {x}) = g.:(A \ {x})
proof
  let x,y be set, g be Function, A be set;
  thus (g +* (x .--> y)).:(A \ {x}) c= g.:(A \ {x})
  proof
    let u be object;
A1: dom(x .--> y) = {x};
    assume u in (g +* (x .--> y)).:(A \ {x});
    then consider z being object such that
A2: z in dom(g +* (x .--> y)) and
A3: z in A \ {x} and
A4: u = (g +* (x .--> y)).z by FUNCT_1:def 6;
A5: not z in {x} by A3,XBOOLE_0:def 5;
    then
A6: z in dom g by A2,A1,FUNCT_4:12;
    u = g.z by A4,A5,A1,FUNCT_4:11;
    hence thesis by A3,A6,FUNCT_1:def 6;
  end;
  let u be object;
  assume u in g.:(A \ {x});
  then consider z being object such that
A7: z in dom g and
A8: z in A \ {x} and
A9: u = g.z by FUNCT_1:def 6;
  not z in {x} by A8,XBOOLE_0:def 5;
  then not z in dom(x .--> y);
  then
A10: u = (g +* (x .--> y)).z by A9,FUNCT_4:11;
  z in dom(g +* (x .--> y)) by A7,FUNCT_4:12;
  hence thesis by A8,A10,FUNCT_1:def 6;
end;
