theorem Th8:
  x@a = x9@a implies x = x9
proof
  assume
A1: x@a = x9@a;
  consider y such that
A2: y@a = x by Def3;
  x = x@(y@a) by A2,Def3
    .= (x@y)@(x9@a) by A1,Th6
    .= x@(x@x9) by A2,Def3;
  then x = x@x9 by Th7;
  hence thesis by Th7;
end;
