reserve A for QC-alphabet;

theorem Th4:
  for x,y being set for g being Function for A being set st not y
in g.:(A \ {x}) holds (g +* (x .--> y)).:(A \ {x}) = (g +* (x .--> y)).:A \ {y}
proof
  let x,y be set, g be Function, A be set;
  assume
A1: not y in g.:(A \ {x});
  thus (g +* (x .--> y)).:(A \ {x}) c= (g +* (x .--> y)).:A \ {y}
  proof
    let u be object;
A2: dom(x .--> y) = {x};
    assume
A3: u in (g +* (x .--> y)).:(A \ {x});
    then consider z being object such that
A4: z in dom(g +* (x .--> y)) and
A5: z in A \ {x} and
A6: u = (g +* (x .--> y)).z by FUNCT_1:def 6;
A7: not z in {x} by A5,XBOOLE_0:def 5;
    then
A8: z in dom g by A4,A2,FUNCT_4:12;
    u = g.z by A6,A7,A2,FUNCT_4:11;
    then u <> y by A1,A5,A8,FUNCT_1:def 6;
    then
A9: not u in {y} by TARSKI:def 1;
    (g +* (x .--> y)).:(A \ {x}) c= (g +* (x .--> y)).: A by RELAT_1:123;
    hence thesis by A3,A9,XBOOLE_0:def 5;
  end;
  let u be object;
  assume
A10: u in (g +* (x .--> y)).:A \ {y};
  then consider z being object such that
A11: z in dom(g +* (x .--> y)) and
A12: z in A and
A13: u = (g +* (x .--> y)).z by FUNCT_1:def 6;
  now
    assume
A14: z in {x};
    then z in dom(x .--> y);
    then u = (x .--> y).z by A13,FUNCT_4:13;
    then u = y by A14,FUNCOP_1:7;
    then u in {y} by TARSKI:def 1;
    hence contradiction by A10,XBOOLE_0:def 5;
  end;
  then z in A \ {x} by A12,XBOOLE_0:def 5;
  hence thesis by A11,A13,FUNCT_1:def 6;
end;
