 Solution

In a two-element first-order structure, $\varphi =\forall x_1(x_0=x_1)$ is false for every first-order variable assignment. Blind substitution of $x_1$ for $x_0$ yields $\forall x_1(x_1=x_1)$, which is true. This does not express the original property evaluated at the value of the inserted term: it has changed a free parameter into a bound variable.

For a direct failure of the substitution axiom, use $\theta (x_0)=\exists x_1(x_0\ne x_1)$. On a two-element domain, $\forall x_0\theta (x_0)$ is true, whereas the captured instance $\exists x_1(x_1\ne x_1)$ is false. Consequently an unrestricted A4 would prove a false implication. The free for substitution condition is necessary for soundness.