theorem Th27:
  p is DTree-yielding & q is DTree-yielding iff p^q is DTree-yielding
proof
A1: rng (p^q) = rng p \/ rng q by FINSEQ_1:31;
  thus
  p is DTree-yielding & q is DTree-yielding implies p^q is DTree-yielding
  by A1,Th9;
  assume
A2: rng (p^q) is constituted-DTrees;
  hence rng p is constituted-DTrees by A1,Th9;
  thus rng q is constituted-DTrees by A1,A2,Th9;
end;
