theorem
  F is associative implies F.:(F[;](x,f),g) = F[;](x,F.:(f,g))
proof
  assume
A1: F is associative;
  per cases;
  suppose
    Y = {};
    hence thesis;
  end;
  suppose
A2: Y <> {};
    now
      let y;
      reconsider x1 = f.y, x2 = g.y as Element of X by A2,FUNCT_2:5;
      thus (F[;](x,F.:(f,g))).y = F.(x,F.:(f,g).y) by A2,Th53
        .= F.(x,F.(x1,x2)) by A2,Th37
        .= F.(F.(x,x1),x2) by A1
        .= F.(F[;](x,f).y,g.y) by A2,Th53;
    end;
    hence thesis by A2,Th38;
  end;
end;
