We consider the following Problem:

  Strict Trs:
    {  h(c(x, y), c(s(z), z), t(w)) ->
       h(z, c(y, x), t(t(c(x, c(y, t(w))))))
     , h(x, c(y, z), t(w)) -> h(c(s(y), x), z, t(c(t(w), w)))
     , h(c(s(x), c(s(0()), y)), z, t(x)) ->
       h(y, c(s(0()), c(x, z)), t(t(c(x, s(x)))))
     , t(t(x)) -> t(c(t(x), x))
     , t(x) -> x
     , t(x) -> c(0(), c(0(), c(0(), c(0(), c(0(), x)))))}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  Arguments of following rules are not normal-forms:
  {  h(c(x, y), c(s(z), z), t(w)) ->
     h(z, c(y, x), t(t(c(x, c(y, t(w))))))
   , h(x, c(y, z), t(w)) -> h(c(s(y), x), z, t(c(t(w), w)))
   , h(c(s(x), c(s(0()), y)), z, t(x)) ->
     h(y, c(s(0()), c(x, z)), t(t(c(x, s(x)))))
   , t(t(x)) -> t(c(t(x), x))}
  
  All above mentioned rules can be savely removed.
  
  We consider the following Problem:
  
    Strict Trs:
      {  t(x) -> x
       , t(x) -> c(0(), c(0(), c(0(), c(0(), c(0(), x)))))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {t(x) -> c(0(), c(0(), c(0(), c(0(), c(0(), x)))))}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(h) = {}, Uargs(c) = {}, Uargs(s) = {}, Uargs(t) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       h(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
                       [0 0]      [0 0]      [0 0]      [0]
       c(x1, x2) = [0 0] x1 + [0 2] x2 + [1]
                   [0 0]      [0 0]      [0]
       s(x1) = [0 0] x1 + [0]
               [0 0]      [0]
       t(x1) = [1 0] x1 + [2]
               [0 0]      [0]
       0() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs: {t(x) -> x}
      Weak Trs: {t(x) -> c(0(), c(0(), c(0(), c(0(), c(0(), x)))))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {t(x) -> x}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(h) = {}, Uargs(c) = {}, Uargs(s) = {}, Uargs(t) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         h(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [0 0] x3 + [0]
                         [0 1]      [0 0]      [0 0]      [0]
         c(x1, x2) = [0 0] x1 + [0 2] x2 + [1]
                     [0 0]      [0 0]      [0]
         s(x1) = [0 0] x1 + [0]
                 [0 0]      [0]
         t(x1) = [1 0] x1 + [2]
                 [0 1]      [0]
         0() = [0]
               [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Weak Trs:
          {  t(x) -> x
           , t(x) -> c(0(), c(0(), c(0(), c(0(), c(0(), x)))))}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(O(1),O(1))
      
      Proof:
        We consider the following Problem:
        
          Weak Trs:
            {  t(x) -> x
             , t(x) -> c(0(), c(0(), c(0(), c(0(), c(0(), x)))))}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(O(1),O(1))
        
        Proof:
          Empty rules are trivially bounded

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