theorem Th134:
  for t,t1 being Tree, xi,nu being Element of t st not xi c= nu &
  not nu c= xi holds (t with-replacement(xi,t1))|nu = t|nu
  proof
    let t,t1 be Tree;
    let xi,nu be Element of t;
    assume Z0: not xi c= nu;
    assume Z1: not nu c= xi;
    let a be FinSequence of NAT;
    hereby
      assume a in (t with-replacement(xi,t1))|nu;
      then reconsider b = a as Element of (t with-replacement(xi,t1))|nu;
      not xi c< nu by Z0,XBOOLE_0:def 8;
      then nu in t with-replacement(xi,t1) by TREES_1:def 9;
      then nu^b in t with-replacement(xi,t1) by TREES_1:def 6;
      then per cases by TREES_1:def 9;
      suppose nu^b in t & not xi c< nu^b;
        hence a in t|nu by TREES_1:def 6;
      end;
      suppose ex r being FinSequence of NAT st r in t1 & nu^b = xi^r;
        then consider r being FinSequence of NAT such that
B1:     r in t1 & nu^b = xi^r;
        nu c= xi^r & xi c= nu^b by B1,Lem8;
        hence a in t|nu by Z0,Z1,Lem8B;
      end;
    end;
    assume a in t|nu;
    then
A2: nu^a in t by TREES_1:def 6;
    not xi c< nu^a by Z0,Z1,Lem8B,XBOOLE_0:def 8;
    then
A3: nu^a in t with-replacement(xi,t1) by A2,TREES_1:def 9;
    not xi c< nu by Z0,XBOOLE_0:def 8;
    then nu in t with-replacement(xi,t1) by TREES_1:def 9;
    hence thesis by A3,TREES_1:def 6;
  end;
