
theorem Th6:
  for T0,T1 being Tree, t being Element of tree(T0,T1) holds (for p
  being Element of T0 st t = <*0*>^p holds t in Leaves tree(T0,T1) iff p in
Leaves T0) & for p being Element of T1 st t = <*1*>^p holds t in Leaves tree(T0
  ,T1) iff p in Leaves T1
proof
  let T0,T1 be Tree, t be Element of tree(T0,T1);
  set RT = tree(T0,T1);
  hereby
    let p be Element of T0;
    assume
A1: t = <*0*>^p;
    hereby
      assume
A2:   t in Leaves RT;
      assume not p in Leaves T0;
      then consider n being Nat such that
A3:   p^<*n*> in T0 by TREES_1:55;
      <*0*>^(p^<*n*>) in RT by A3,TREES_3:69;
      then (<*0*>^p)^<*n*> in RT by FINSEQ_1:32;
      hence contradiction by A1,A2,TREES_1:55;
    end;
    assume
A4: p in Leaves T0;
    assume not t in Leaves RT;
    then consider n being Nat such that
A5: t^<*n*> in RT by TREES_1:55;
    <*0*>^(p^<*n*>) in RT by A1,A5,FINSEQ_1:32;
    then p^<*n*> in T0 by TREES_3:69;
    hence contradiction by A4,TREES_1:55;
  end;
  let p be Element of T1;
  assume
A6: t = <*1*>^p;
  hereby
    assume
A7: t in Leaves RT;
    assume not p in Leaves T1;
    then consider n being Nat such that
A8: p^<*n*> in T1 by TREES_1:55;
    <*1*>^(p^<*n*>) in RT by A8,TREES_3:70;
    then (<*1*>^p)^<*n*> in RT by FINSEQ_1:32;
    hence contradiction by A6,A7,TREES_1:55;
  end;
  assume
A9: p in Leaves T1;
  assume not t in Leaves RT;
  then consider n being Nat such that
A10: t^<*n*> in RT by TREES_1:55;
  <*1*>^(p^<*n*>) in RT by A6,A10,FINSEQ_1:32;
  then p^<*n*> in T1 by TREES_3:70;
  hence contradiction by A9,TREES_1:55;
end;
