Problem:
 minus(minus(x)) -> x
 minus(h(x)) -> h(minus(x))
 minus(f(x,y)) -> f(minus(y),minus(x))

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {3}
   transitions:
    f1(4,4) -> 4,3
    minus1(2) -> 4*
    minus1(1) -> 4*
    h1(4) -> 4,3
    minus0(2) -> 3*
    minus0(1) -> 3*
    h0(2) -> 1*
    h0(1) -> 1*
    f0(1,2) -> 2*
    f0(2,1) -> 2*
    f0(1,1) -> 2*
    f0(2,2) -> 2*
  problem:
   
  Qed