From (1)-(2) we get $y z(x-t)=x-t$,
thus, we have $y z=1$ or $x=t$.
If $y z=1$, then from (1), we get $x=x+y+z$, so $y+z=0$. However, the system of equations $y \cdot z=1$ and $y+z=0$ has no real solutions, so it must be that $x=t$.
Substituting $x=t$ into (3), we get $z=\frac{2 x}{x^{2}-1}$.
Substituting $x=t$ into (4), we get $y=\frac{2 x}{x^{2}-1}$.
Substituting (5) and (6) into (1), we get $\frac{4 x^{3}}{\left(x^{2}-1\right)^{2}}=x+\frac{4 x}{x^{2}-1}$.
$x=0$ is clearly a solution to the above equation. When $x \neq 0$, the above equation can be transformed into
$x^{4}-2 x^{2}-3=0$, solving this yields the real roots $x= \pm \sqrt{3}$.
Thus, $t=0$ or $t= \pm \sqrt{3}$.
Similarly, we get $y=z=0$ or $y=z= \pm \sqrt{3}$.
Therefore, the solutions $(x, y, z, t)$ to the original system of equations are:
$$
(0,0,0,0),(\sqrt{3}, \sqrt{3}, \sqrt{3}, \sqrt{3}),(-\sqrt{3},-\sqrt{3},-\sqrt{3},-\sqrt{3}) \text {. }
$$
Example 20 (13th All-Russian Mathematical Olympiad) Solve the system of equations in the set of real numbers:
$$
\left\{\begin{array}{l}
x y z=x+y+z, \\
y z t=y+z+t, \\
z t x=z+t+x, \\
t x y=t+x+y .
\end{array}\right.
$$