We consider the following Problem:

  Strict Trs:
    {  c(c(c(b(x)))) -> a(1(), b(c(x)))
     , b(c(b(c(x)))) -> a(0(), a(1(), x))
     , a(0(), x) -> c(c(x))
     , a(1(), x) -> c(b(x))}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  c(c(c(b(x)))) -> a(1(), b(c(x)))
       , b(c(b(c(x)))) -> a(0(), a(1(), x))
       , a(0(), x) -> c(c(x))
       , a(1(), x) -> c(b(x))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  a(0(), x) -> c(c(x))
       , a(1(), x) -> c(b(x))}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(c) = {1}, Uargs(b) = {1}, Uargs(a) = {2}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       c(x1) = [1 0] x1 + [0]
               [0 0]      [1]
       b(x1) = [1 0] x1 + [0]
               [0 0]      [1]
       a(x1, x2) = [0 0] x1 + [1 0] x2 + [1]
                   [0 0]      [0 0]      [1]
       1() = [0]
             [0]
       0() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  c(c(c(b(x)))) -> a(1(), b(c(x)))
         , b(c(b(c(x)))) -> a(0(), a(1(), x))}
      Weak Trs:
        {  a(0(), x) -> c(c(x))
         , a(1(), x) -> c(b(x))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {b(c(b(c(x)))) -> a(0(), a(1(), x))}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(c) = {1}, Uargs(b) = {1}, Uargs(a) = {2}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         c(x1) = [1 0] x1 + [0]
                 [0 0]      [1]
         b(x1) = [1 0] x1 + [1]
                 [0 0]      [1]
         a(x1, x2) = [0 1] x1 + [1 0] x2 + [0]
                     [0 0]      [0 0]      [1]
         1() = [0]
               [1]
         0() = [0]
               [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs: {c(c(c(b(x)))) -> a(1(), b(c(x)))}
        Weak Trs:
          {  b(c(b(c(x)))) -> a(0(), a(1(), x))
           , a(0(), x) -> c(c(x))
           , a(1(), x) -> c(b(x))}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        We consider the following Problem:
        
          Strict Trs: {c(c(c(b(x)))) -> a(1(), b(c(x)))}
          Weak Trs:
            {  b(c(b(c(x)))) -> a(0(), a(1(), x))
             , a(0(), x) -> c(c(x))
             , a(1(), x) -> c(b(x))}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The problem is match-bounded by 0.
          The enriched problem is compatible with the following automaton:
          {  c_0(1) -> 3
           , c_0(2) -> 3
           , c_0(4) -> 1
           , c_0(5) -> 1
           , b_0(4) -> 2
           , b_0(5) -> 2
           , a_0(4, 4) -> 3
           , a_0(4, 5) -> 3
           , a_0(5, 4) -> 3
           , a_0(5, 5) -> 3
           , 1_0() -> 4
           , 0_0() -> 5}

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