YES

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

Proof:
 DP Processor:
  DPs:
   .#(.(x,y),z) -> .#(y,z)
   .#(.(x,y),z) -> .#(x,.(y,z))
  TRS:
   .(.(x,y),z) -> .(x,.(y,z))
  KBO Processor:
   argument filtering:
    pi(.) = [0,1]
    pi(.#) = 0
   usable rules:
    
   weight function:
    w0 = 1
    w(.#) = w(.) = 1
   precedence:
    .# ~ .
   problem:
    DPs:
     
    TRS:
     .(.(x,y),z) -> .(x,.(y,z))
   Qed