Let $x = b - a$, $y = c - b$, $z = d - c$. Then, $x, y, z$ are integers satisfying $0 < x < y < z$. Furthermore, since $x + y + z = d - a \le 9 - 1 = 8$, we see that the triple $(x, y, z)$ must be one of the following $4$ triples:
$$(1, 2, 3),\ (1, 2, 4),\ (1, 2, 5),\ (1, 3, 4).$$
When $(x, y, z) = (1, 2, 3)$, we get $(a, b, c, d) = (a, a+1, a+3, a+6)$. So, corresponding to the cases $a = 1, 2, 3$, we get $3$ quadruples $(a, b, c, d)$ satisfying the required condition. Similarly, when $(x, y, z) = (1, 2, 4)$, we get $2$ quadruples $(a, b, c, d)$ satisfying the requirement corresponding to the cases $a = 1, 2$. We also find if $(x, y, z) = (1, 2, 5)$ or if $(x, y, z) = (1, 3, 4)$ only $1$ quadruple $(a, b, c, d)$ corresponding to $a = 1$ will satisfy the requirement. Therefore, there are only $3 + 2 + 1 + 1 = 7$ quadruples $(a, b, c, d)$ satisfying the requirement.