We consider the following Problem:

  Strict Trs:
    {  minus(minus(x)) -> x
     , minus(h(x)) -> h(minus(x))
     , minus(f(x, y)) -> f(minus(y), minus(x))}
  StartTerms: basic terms
  Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  minus(minus(x)) -> x
       , minus(h(x)) -> h(minus(x))
       , minus(f(x, y)) -> f(minus(y), minus(x))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {minus(minus(x)) -> x}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(minus) = {}, Uargs(h) = {1}, Uargs(f) = {1, 2}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       minus(x1) = [1 0] x1 + [2]
                   [0 1]      [1]
       h(x1) = [1 0] x1 + [0]
               [0 1]      [0]
       f(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                   [0 0]      [1 1]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  minus(h(x)) -> h(minus(x))
         , minus(f(x, y)) -> f(minus(y), minus(x))}
      Weak Trs: {minus(minus(x)) -> x}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      We consider the following Problem:
      
        Strict Trs:
          {  minus(h(x)) -> h(minus(x))
           , minus(f(x, y)) -> f(minus(y), minus(x))}
        Weak Trs: {minus(minus(x)) -> x}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The problem is match-bounded by 1.
        The enriched problem is compatible with the following automaton:
        {  minus_0(2) -> 1
         , minus_1(2) -> 3
         , h_0(2) -> 2
         , h_1(3) -> 1
         , h_1(3) -> 3
         , f_0(2, 2) -> 2
         , f_1(3, 3) -> 1
         , f_1(3, 3) -> 3}

Hurray, we answered YES(?,O(n^1))