theorem Th25:
  p is Tree-yielding & q is Tree-yielding iff p^q is Tree-yielding
proof
A1: rng (p^q) = rng p \/ rng q by FINSEQ_1:31;
  thus
  p is Tree-yielding & q is Tree-yielding implies p^q is Tree-yielding
  by A1,Th3;
  assume
A2: rng (p^q) is constituted-Trees;
  hence rng p is constituted-Trees by A1,Th3;
  thus rng q is constituted-Trees by A1,A2,Th3;
end;
