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:
 Bounds Processor:
  bound: 0
  enrichment: roof
  automaton:
   final states: {2,1}
   transitions:
    f30() -> 1*
    10() -> 2*
  problem:
   
  Qed