YES

Problem:
 -(-(neg(x),neg(x)),-(neg(y),neg(y))) -> -(-(x,y),-(x,y))

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {2,1}
   transitions:
    -0(1,2) -> 1*
    -0(2,1) -> 1*
    -0(1,1) -> 1*
    -0(2,2) -> 1*
    neg0(2) -> 2*
    neg0(1) -> 2*
    -1(1,2) -> 1*
    -1(2,1) -> 1*
    -1(1,1) -> 1*
    -1(2,2) -> 1*
  problem:
   
  Qed