theorem Th18:
  a,b @@ a9,b9 & b,c @@ b9,c9 implies a,c @@ a9,c9
proof
  assume
A1: a,b @@ a9,b9;
  assume
A2: b,c @@ b9,c9;
  (b9@b)@(a@c9) = (a@b9)@(b@c9) by Def3
    .= (b@a9)@(b@c9) by A1
    .= (c@b9)@(b@a9) by A2
    .= (b9@b)@(c@a9) by Def3;
  hence a@c9 = c@a9 by Th8;
end;
