YES

Problem:
 .(1(),x) -> x
 .(x,1()) -> x
 .(i(x),x) -> 1()
 .(x,i(x)) -> 1()
 i(1()) -> 1()
 i(i(x)) -> x
 .(i(y),.(y,z)) -> z
 .(y,.(i(y),z)) -> z

Proof:
 DP Processor:
  DPs:
   
  TRS:
   .(1(),x) -> x
   .(x,1()) -> x
   .(i(x),x) -> 1()
   .(x,i(x)) -> 1()
   i(1()) -> 1()
   i(i(x)) -> x
   .(i(y),.(y,z)) -> z
   .(y,.(i(y),z)) -> z
  Qed