YES

Problem:
 ++(nil(),y) -> y
 ++(x,nil()) -> x
 ++(.(x,y),z) -> .(x,++(y,z))
 ++(++(x,y),z) -> ++(x,++(y,z))

Proof:
 DP Processor:
  DPs:
   ++#(.(x,y),z) -> ++#(y,z)
   ++#(++(x,y),z) -> ++#(y,z)
   ++#(++(x,y),z) -> ++#(x,++(y,z))
  TRS:
   ++(nil(),y) -> y
   ++(x,nil()) -> x
   ++(.(x,y),z) -> .(x,++(y,z))
   ++(++(x,y),z) -> ++(x,++(y,z))
  Subterm Criterion Processor:
   simple projection:
    pi(++#) = 0
   problem:
    DPs:
     
    TRS:
     ++(nil(),y) -> y
     ++(x,nil()) -> x
     ++(.(x,y),z) -> .(x,++(y,z))
     ++(++(x,y),z) -> ++(x,++(y,z))
   Qed