It is clear that inequality (16) contradicts (10). On the other hand, integrating inequality (14) from n to n + 1, we get Since z ( t ) is nonincreasing for t > T, from the above inequality, we obtain that and so, we have In view of (15), (8), and (9), we obtain from the above inequality that and Since n ≤ s < n + 1 , (17) contradicts (11). Case 2 . τ ≤ 1 . Then z ( t − τ ) ≤ z ([ t − 1]) for n + 1 > t ≥ n > T, and from (13), we obtain that 9 0 = z $^{′}$( t ) + B ( t ) z ( t − τ $_{) exp}$  t ∫ t − τ a ( s ) $_{ds}$  + C ( t ) z ([ t − $_{1]) exp}$   t ∫ [ t − 1] a ( s ) $_{ds}$   ≥ z $^{′}$( t ) + z ( t − τ ) P ( t ) , (18) n +1 ∫ n ∏ [ s − 1] <t$_{j}$ ≤ s (1 − b$_{j}$ ) c ( s $_{) exp}$   s ∫ [ s − 1] a ( u ) $_{du}$   ds ≤ 1 . (17) n +1 ∫ n ∏ s − τ<t$_{j}$ ≤ s (1 − b$_{j}$ ) b ( s $_{) exp}$  s ∫ s − τ a ( u ) $_{du}$  ds ≤ 1 , n +1 ∫ n P ( s ) ds ≤ 1 . z ( n + 1) + z ( n $_{)}$  n +1 ∫ n P ( s ) ds − $_{1}$  ≤ 0 z ( n + 1) − z ( n ) + n +1 ∫ n P ( s ) z ( s − 1) ds ≤ 0 .