theorem Th17:
  x,y @@ a,b & x,y @@ c,d implies a,b @@ c,d
proof
  assume
A1: x,y @@ a,b;
  assume
A2: x,y @@ c,d;
  (y@x)@(a@d) = (y@a)@(x@d) by Def3
    .= (x@b)@(x@d) by A1
    .= (x@b)@(y@c) by A2
    .= (y@x)@(b@c) by Def3;
  hence a@d = b@c by Th8;
end;
