theorem
  for f be Function, n,m be object holds (f +* (n .--> m) +* (m .--> n)).n=
  m
proof
  let f be Function,n,m be object;
  set mn=m .--> n, nm=n .--> m;
A1: m in dom mn by TARSKI:def 1;
  per cases;
  suppose
A2: n=m;
    hence (f +* nm +* mn).n=mn.m by A1,Th13
      .=m by A2,FUNCOP_1:72;
  end;
  suppose
A3: n<>m;
A4: n in dom nm by TARSKI:def 1;
    not n in dom mn by A3,TARSKI:def 1;
    hence (f +* nm +* mn).n=(f +* nm).n by Th11
      .=nm.n by A4,Th13
      .=m by FUNCOP_1:72;
  end;
end;
