theorem Th39:
  'not' A <> (#)B & 'not' A <> B '&' C
proof
  set e2 = elementary_tree 2;
  set e1 = elementary_tree 1;
  set E = e1 --> [1,0];
  set F = e1 --> [1,1];
  reconsider e = {} as Element of dom 'not' A by TREES_1:22;
A1: {} in dom (#)B & not ex u st u in dom B & e = <*0*>^u & ((#)B).e = B.u
  by TREES_1:22;
A2: <*0*> in e1 by TARSKI:def 2,TREES_1:51;
  then
A3: <*0*> in dom E by FUNCOP_1:13;
  then
A4: dom 'not' A = dom E with-replacement (<*0*>,dom A) by TREES_2:def 11;
A5: <*0*> in dom F by A2,FUNCOP_1:13;
  then dom (#)B = dom F with-replacement (<*0*>,dom B) by TREES_2:def 11;
  then
A6: ((#) B).e = F.e by A5,A1,TREES_2:def 11;
  e in e1 by TREES_1:22;
  then
A7: E.e = [1,0] & F.e = [1,1] by FUNCOP_1:7;
  ( not ex u st u in dom A & e = <*0*>^u & ('not' A).e = A.u)& [1,0] <> [
  1,1] by XTUPLE_0:1;
  hence 'not' A <> (#)B by A3,A4,A6,A7,TREES_2:def 11;
  set y = (e2-->[2,0]) with-replacement (<*0*>,B);
A8: <*1*> in e2 & not <*0*> is_a_proper_prefix_of <*1*> by TREES_1:28,52;
A9: <*0*> in e2 & dom (e2 --> [2,0]) = e2 by FUNCOP_1:13,TREES_1:28;
  then dom y = dom(e2-->[2,0]) with-replacement (<*0*>,dom B) by TREES_2:def 11
;
  then
A10: <*1*> in dom y by A9,A8,TREES_1:def 9;
  then dom (B '&' C) = dom y with-replacement (<*1*>,dom C) by TREES_2:def 11;
  then
A11: <*1*> in dom (B '&' C) by A10,TREES_1:def 9;
A12: now
    assume <*1*> in dom E;
    then <*1*> = {} or <*1*> = <*0*> by TARSKI:def 2,TREES_1:51;
    hence contradiction by TREES_1:3;
  end;
  assume not thesis;
  then ex s st s in dom A & <*1*> = <*0*>^s by A3,A4,A11,A12,TREES_1:def 9;
  then <*0*> is_a_prefix_of <*1*> by TREES_1:1;
  hence contradiction by TREES_1:3;
end;
