Let us set the system for the possible usage in the remaining part of the paper. Consider a solution y ( t, 0 , x$_{0}$ ) , x$_{0}$ ∈ ˜ Γ , of (2.2). Denote the first meeting point of the solution with the surface Γ , provided the point exists, by y ( ζ, 0 , x$_{0}$ ) . The following conditions are sufficient for the continuation property. To verify the continuation of the solutions of (2.1), the following theorems can be applied. (C9) (a) Every solution y ( t, 0 , x$_{0}$ ) , x$_{0}$ ∈ D, of (2.2) is continuable to either ∞ or Γ as time increases, (b) Every solution y ( t, 0 , x$_{0}$ ) , x$_{0}$ ∈ D, of (2.2) is continuable to either −∞ or ˜ Γ as time decreases. (C10) For all x$_{0}$ ∈ D, the solution y ( t, 0 , x$_{0}$ ) of (2.2) does not intersect ˜ Γ before it meets the surface Γ as time increases. Theorem 2.1 [1] Assume that conditions ( C 8) and ( C 9) are valid. Then, every solution x ( t ) = x ( t, 0 , x$_{0}$ ) , x$_{0}$ ∈ D of (2.1) is continuable on $^{R}$. Now, we will present a condition which is sufficient for the group property . In other words, for each x$_{0}$ ∈ D and a positive number s such that y ( s, 0 , x$_{0}$ ) ∈ ˜ Γ , there exists a number r, 0 ≤ r < s, such that y ( r, 0 , x$_{0}$ ) ∈ Γ . It is easy to verify that the condition ( C 10) is equivalent to the assertion that for all x$_{0}$ ∈ D, the solution y ( t, 0 , x$_{0}$ ) of (2.2) does not intersect Γ before it meets the surface ˜ Γ as time decreases. In other words, for each x$_{0}$ ∈ D and a negative number s such that y ( s, 0 , x$_{0}$ ) ∈ ˜ Γ , there exists a number r, s < r ≤ 0 , such that y ( r, 0 , x$_{0}$ ) ∈ ˜ Γ . Theorem 2.2 (The group property) Assume that conditions (C1)-(C10) are valid. Then, x ( t$_{2}$, 0 , x ( t$_{1}$, 0 , x$_{0}$ )) = x ( t$_{2}$ + t$_{1}$, 0 , x$_{0}$ ) , for all t$_{1}$, t$_{2}$ ∈ $^{R}$. Proof. Denote by ξ ( t ) = x ( t + ¯ t ) , for a fixed ¯ t ∈ $^{R}$. It can be verified that the sequence { θ$_{i}$ − ¯ t } is a set of discontinuity moments of ξ ( t ) and the function is a solution of (2.1) [1]. The next step is to show that the following equality x ( − t, 0 , x ( t, 0 , x$_{0}$ )) = x$_{0}$, holds for all x$_{0}$ ∈ D and t ∈ $^{R}$. Consider the case t > 0 . If the set of discontinuity moments { θ$_{i}$ } is empty, the proof is same with that for continuous dynamical systems [23]. Because of the condition ( C 2) , which corresponds to invertibility of the jump function J, the equality x ( θ$_{i}$, 0 , x ( θ$_{i}$ +)) = x ( θ$_{i}$ ) , holds for all i ∈ A . Assuming that $^{θ}$− 1 < 0 < θ$_{1}$, we should verify x ( − θ$_{1}$, 0 , x ( θ$_{1}$, 0 , x$_{0}$ )) = x$_{0}$. Denote by ¯ x ( t ) = x ( t, 0 , x ( θ$_{1}$ )) . The point x ( θ$_{1}$ ) lies on the discontinuity surface Γ . By condition ( C 3) the solution ¯ x ( t ) is a trajectory of y ′ = f ( y ) for decreasing t. Condition ( C 10) , part ( a ) , implies that the trajectory ¯ x ( t ) cannot meet with ˜ Γ if t > − θ$_{1}$ as time decreases. That is, ¯ x ( − θ$_{1}$ ) = x$_{0}$ as the dynamics is continuous. The proof for t < 0 can be done in a similar way. □ y ′ = f ( y ) (2.2) 5