reserve X,Y,Z,X1,X2,Y1,Y2 for set, x,y,z,t,x1,x2 for object,
  f,g,h,f1,f2,g1,g2 for Function;

theorem Th52:
  Funcs(X,{x}) = {X --> x}
proof
  thus Funcs(X,{x}) c= {X --> x}
  proof
    let y be object;
    assume y in Funcs(X,{x});
    then consider f such that
A1: y = f and
A2: dom f = X and
A3: rng f c= {x} by FUNCT_2:def 2;
A4: now
      set z = the Element of X;
A5:   X <> {} implies z in X;
      assume for z holds not z in X;
      hence f = X --> x by A2,A5;
    end;
    now
      given z such that
A6:   z in X;
      f.z in rng f by A2,A6,FUNCT_1:def 3;
      then f.z = x & {f.z} c= rng f by A3,TARSKI:def 1;
      then rng f = {x} by A3;
      hence f = X --> x by A2,FUNCOP_1:9;
    end;
    hence thesis by A1,A4,TARSKI:def 1;
  end;
  let y be object;
  assume y in {X --> x};
  then
A7: y = X --> x by TARSKI:def 1;
  dom(X --> x) = X & rng(X --> x) c= {x} by FUNCOP_1:13;
  hence thesis by A7,FUNCT_2:def 2;
end;
