We consider the following Problem:

  Strict Trs:
    {  f(f(x)) -> f(c(f(x)))
     , f(f(x)) -> f(d(f(x)))
     , g(c(x)) -> x
     , g(d(x)) -> x
     , g(c(h(0()))) -> g(d(1()))
     , g(c(1())) -> g(d(h(0())))
     , g(h(x)) -> g(x)}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(f(x)) -> f(c(f(x)))
       , f(f(x)) -> f(d(f(x)))
       , g(c(x)) -> x
       , g(d(x)) -> x
       , g(c(h(0()))) -> g(d(1()))
       , g(c(1())) -> g(d(h(0())))
       , g(h(x)) -> g(x)}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {g(h(x)) -> g(x)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {}, Uargs(c) = {}, Uargs(d) = {}, Uargs(g) = {},
        Uargs(h) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1) = [1 0] x1 + [0]
               [0 0]      [1]
       c(x1) = [0 0] x1 + [0]
               [1 0]      [0]
       d(x1) = [0 0] x1 + [0]
               [1 0]      [0]
       g(x1) = [0 1] x1 + [1]
               [0 0]      [1]
       h(x1) = [0 0] x1 + [0]
               [0 1]      [2]
       0() = [0]
             [0]
       1() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  f(f(x)) -> f(c(f(x)))
         , f(f(x)) -> f(d(f(x)))
         , g(c(x)) -> x
         , g(d(x)) -> x
         , g(c(h(0()))) -> g(d(1()))
         , g(c(1())) -> g(d(h(0())))}
      Weak Trs: {g(h(x)) -> g(x)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {g(c(1())) -> g(d(h(0())))}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(f) = {}, Uargs(c) = {}, Uargs(d) = {}, Uargs(g) = {},
          Uargs(h) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         f(x1) = [1 0] x1 + [0]
                 [0 0]      [1]
         c(x1) = [0 0] x1 + [0]
                 [1 0]      [0]
         d(x1) = [0 0] x1 + [0]
                 [1 0]      [0]
         g(x1) = [0 1] x1 + [1]
                 [0 0]      [1]
         h(x1) = [0 0] x1 + [0]
                 [0 1]      [0]
         0() = [0]
               [0]
         1() = [2]
               [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  f(f(x)) -> f(c(f(x)))
           , f(f(x)) -> f(d(f(x)))
           , g(c(x)) -> x
           , g(d(x)) -> x
           , g(c(h(0()))) -> g(d(1()))}
        Weak Trs:
          {  g(c(1())) -> g(d(h(0())))
           , g(h(x)) -> g(x)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component:
          {  f(f(x)) -> f(c(f(x)))
           , f(f(x)) -> f(d(f(x)))}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(f) = {}, Uargs(c) = {}, Uargs(d) = {}, Uargs(g) = {},
            Uargs(h) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           f(x1) = [0 1] x1 + [0]
                   [0 0]      [1]
           c(x1) = [1 2] x1 + [0]
                   [0 0]      [0]
           d(x1) = [1 0] x1 + [0]
                   [0 0]      [0]
           g(x1) = [1 0] x1 + [0]
                   [0 0]      [0]
           h(x1) = [1 0] x1 + [0]
                   [0 0]      [0]
           0() = [0]
                 [0]
           1() = [0]
                 [2]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  g(c(x)) -> x
             , g(d(x)) -> x
             , g(c(h(0()))) -> g(d(1()))}
          Weak Trs:
            {  f(f(x)) -> f(c(f(x)))
             , f(f(x)) -> f(d(f(x)))
             , g(c(1())) -> g(d(h(0())))
             , g(h(x)) -> g(x)}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The weightgap principle applies, where following rules are oriented strictly:
          
          TRS Component: {g(c(h(0()))) -> g(d(1()))}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(f) = {}, Uargs(c) = {}, Uargs(d) = {}, Uargs(g) = {},
              Uargs(h) = {}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             f(x1) = [0 0] x1 + [0]
                     [1 1]      [2]
             c(x1) = [1 0] x1 + [1]
                     [0 0]      [0]
             d(x1) = [1 0] x1 + [0]
                     [1 0]      [0]
             g(x1) = [1 0] x1 + [0]
                     [1 0]      [3]
             h(x1) = [1 0] x1 + [0]
                     [0 0]      [0]
             0() = [0]
                   [0]
             1() = [0]
                   [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Strict Trs:
              {  g(c(x)) -> x
               , g(d(x)) -> x}
            Weak Trs:
              {  g(c(h(0()))) -> g(d(1()))
               , f(f(x)) -> f(c(f(x)))
               , f(f(x)) -> f(d(f(x)))
               , g(c(1())) -> g(d(h(0())))
               , g(h(x)) -> g(x)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            The weightgap principle applies, where following rules are oriented strictly:
            
            TRS Component: {g(c(x)) -> x}
            
            Interpretation of nonconstant growth:
            -------------------------------------
              The following argument positions are usable:
                Uargs(f) = {}, Uargs(c) = {}, Uargs(d) = {}, Uargs(g) = {},
                Uargs(h) = {}
              We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
              Interpretation Functions:
               f(x1) = [1 0] x1 + [0]
                       [0 1]      [0]
               c(x1) = [1 0] x1 + [0]
                       [0 1]      [0]
               d(x1) = [1 0] x1 + [0]
                       [0 0]      [0]
               g(x1) = [1 0] x1 + [1]
                       [0 1]      [1]
               h(x1) = [1 0] x1 + [0]
                       [0 1]      [0]
               0() = [0]
                     [0]
               1() = [0]
                     [0]
            
            The strictly oriented rules are moved into the weak component.
            
            We consider the following Problem:
            
              Strict Trs: {g(d(x)) -> x}
              Weak Trs:
                {  g(c(x)) -> x
                 , g(c(h(0()))) -> g(d(1()))
                 , f(f(x)) -> f(c(f(x)))
                 , f(f(x)) -> f(d(f(x)))
                 , g(c(1())) -> g(d(h(0())))
                 , g(h(x)) -> g(x)}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^1))
            
            Proof:
              The weightgap principle applies, where following rules are oriented strictly:
              
              TRS Component: {g(d(x)) -> x}
              
              Interpretation of nonconstant growth:
              -------------------------------------
                The following argument positions are usable:
                  Uargs(f) = {}, Uargs(c) = {}, Uargs(d) = {}, Uargs(g) = {},
                  Uargs(h) = {}
                We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                Interpretation Functions:
                 f(x1) = [1 0] x1 + [0]
                         [0 1]      [0]
                 c(x1) = [1 0] x1 + [0]
                         [0 1]      [0]
                 d(x1) = [1 0] x1 + [0]
                         [0 1]      [0]
                 g(x1) = [1 0] x1 + [1]
                         [0 1]      [1]
                 h(x1) = [1 0] x1 + [0]
                         [0 1]      [0]
                 0() = [0]
                       [0]
                 1() = [0]
                       [0]
              
              The strictly oriented rules are moved into the weak component.
              
              We consider the following Problem:
              
                Weak Trs:
                  {  g(d(x)) -> x
                   , g(c(x)) -> x
                   , g(c(h(0()))) -> g(d(1()))
                   , f(f(x)) -> f(c(f(x)))
                   , f(f(x)) -> f(d(f(x)))
                   , g(c(1())) -> g(d(h(0())))
                   , g(h(x)) -> g(x)}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(O(1),O(1))
              
              Proof:
                We consider the following Problem:
                
                  Weak Trs:
                    {  g(d(x)) -> x
                     , g(c(x)) -> x
                     , g(c(h(0()))) -> g(d(1()))
                     , f(f(x)) -> f(c(f(x)))
                     , f(f(x)) -> f(d(f(x)))
                     , g(c(1())) -> g(d(h(0())))
                     , g(h(x)) -> g(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))