theorem Th133:
  for t,t1 being DecoratedTree, xi,nu being Node 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 DecoratedTree;
    let xi,nu be Node of t;
    assume Z0: not xi c= nu;
    assume Z1: not nu c= xi;
    not xi c< nu by Z0,XBOOLE_0:def 8;
    then
A1: nu in dom t with-replacement(xi,dom t1)
    = dom(t with-replacement(xi,t1)) by TREES_1:def 9,TREES_2:def 11;
A2: dom((t with-replacement(xi,t1))|nu) = (dom(t with-replacement(xi,t1)))|nu
    by TREES_2:def 10;
    hence
A4: dom((t with-replacement(xi,t1))|nu) = (dom t)|nu by Z0,Z1,A1,Th134
    .= dom (t|nu) by TREES_2:def 10;
    let p be Node of (t with-replacement(xi,t1))|nu;
A8: (dom t)|nu = dom (t|nu) by TREES_2:def 10;
    then
A5: nu^p in dom t & (t|nu).p = t.(nu^p) by A4,TREES_1:def 6,TREES_2:def 10;
    not xi c< nu^p by Z0,Z1,Lem8B,XBOOLE_0:def 8;
    then
A7: nu^p in dom t with-replacement(xi,dom t1) by A5,TREES_1:def 9;
A6: not ex r being FinSequence of NAT st r in dom t1 & nu^p = xi^r &
    (t with-replacement(xi,t1)).(nu^p) = t1.r by Z0,Z1,Lem8B,TREES_1:1;
    thus ((t with-replacement(xi,t1))|nu).p
    = (t with-replacement(xi,t1)).(nu^p) by A2,TREES_2:def 10
    .= t.(nu^p) by A6,A7,TREES_2:def 11
    .= (t|nu).p by A8,TREES_2:def 10,A4;
  end;
