theorem Th66:
  |- f^<*All(x,'not' 'not' p)*> implies |- f^<*All(x,p)*>
proof
  assume
A1: |- f^<*All(x,'not' 'not' p)*>;
  consider y0 such that
A2: not y0 in still_not-bound_in (f^<*All(x,p)*>) by Th64;
  Ant(f^<*All(x,'not' 'not' p)*>) = f & Suc(f^<*All(x,'not' 'not' p)*>) =
  All( x,'not' 'not' p) by Th5;
  then |- f^<*('not' 'not' p).(x,y0)*> by A1,Th42;
  then |- f^<*'not' (('not' p).(x,y0))*> by Th56;
  then
A3: |- f^<*'not' 'not' (p.(x,y0))*> by Th56;
  set f1 = f^<*p.(x,y0)*>;
A4: not y0 in still_not-bound_in f \/ still_not-bound_in <*All(x,p)*> by A2
,Th58;
  then not y0 in still_not-bound_in f by XBOOLE_0:def 3;
  then
A5: not y0 in still_not-bound_in Ant(f1) by Th5;
  not y0 in still_not-bound_in <*All(x,p)*> by A4,XBOOLE_0:def 3;
  then
A6: not y0 in still_not-bound_in All(x,p) by Th59;
  Suc(f1) = p.(x,y0) by Th5;
  then |- Ant(f1)^<* All(x,p)*> by A3,A5,A6,Th43,Th54;
  hence thesis by Th5;
end;
