We consider the following Problem:

  Strict Trs:
    {  f1() -> g1()
     , f1() -> g2()
     , f2() -> g1()
     , f2() -> g2()
     , g1() -> h1()
     , g1() -> h2()
     , g2() -> h1()
     , g2() -> h2()
     , e1(h1(), h2(), x, y, z) -> e2(x, x, y, z, z)
     , e2(f1(), x, y, z, f2()) -> e3(x, y, x, y, y, z, y, z, x, y, z)
     , e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) ->
       e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
     , e4(g1(), x1, g2(), x1, g1(), x1, g2(), x1, x, y, z) ->
       e1(x1, x1, x, y, z)}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  Arguments of following rules are not normal-forms:
  {  e4(g1(), x1, g2(), x1, g1(), x1, g2(), x1, x, y, z) ->
     e1(x1, x1, x, y, z)
   , e2(f1(), x, y, z, f2()) -> e3(x, y, x, y, y, z, y, z, x, y, z)}
  
  All above mentioned rules can be savely removed.
  
  We consider the following Problem:
  
    Strict Trs:
      {  f1() -> g1()
       , f1() -> g2()
       , f2() -> g1()
       , f2() -> g2()
       , g1() -> h1()
       , g1() -> h2()
       , g2() -> h1()
       , g2() -> h2()
       , e1(h1(), h2(), x, y, z) -> e2(x, x, y, z, z)
       , e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) ->
         e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  f1() -> g1()
       , f1() -> g2()}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(e1) = {}, Uargs(e2) = {}, Uargs(e3) = {}, Uargs(e4) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f1() = [2]
              [0]
       g1() = [0]
              [0]
       g2() = [0]
              [0]
       f2() = [0]
              [0]
       h1() = [0]
              [0]
       h2() = [0]
              [0]
       e1(x1, x2, x3, x4, x5) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0 0] x4 + [1 1] x5 + [1]
                                [0 0]      [0 0]      [0 0]      [0 0]      [0 0]      [1]
       e2(x1, x2, x3, x4, x5) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [1 1] x4 + [0 0] x5 + [0]
                                [1 0]      [0 0]      [0 0]      [0 0]      [1 0]      [0]
       e3(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11) = [1 1] x1 + [0 0] x2 + [1 1] x3 + [0 0] x4 + [1 1] x5 + [0 0] x6 + [1 1] x7 + [0 0] x8 + [0 0] x9 + [0 0] x10 + [0 0] x11 + [0]
                                                          [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [0 0]       [0 0]       [0]
       e4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11) = [1 1] x1 + [0 0] x2 + [1 1] x3 + [0 0] x4 + [1 1] x5 + [0 0] x6 + [1 0] x7 + [0 1] x8 + [0 0] x9 + [0 0] x10 + [0 0] x11 + [0]
                                                          [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [0 0]       [0 0]       [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  f2() -> g1()
         , f2() -> g2()
         , g1() -> h1()
         , g1() -> h2()
         , g2() -> h1()
         , g2() -> h2()
         , e1(h1(), h2(), x, y, z) -> e2(x, x, y, z, z)
         , e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) ->
           e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)}
      Weak Trs:
        {  f1() -> g1()
         , f1() -> g2()}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component:
        {  f2() -> g1()
         , f2() -> g2()}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(e1) = {}, Uargs(e2) = {}, Uargs(e3) = {}, Uargs(e4) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         f1() = [0]
                [0]
         g1() = [0]
                [0]
         g2() = [0]
                [0]
         f2() = [2]
                [0]
         h1() = [0]
                [0]
         h2() = [0]
                [0]
         e1(x1, x2, x3, x4, x5) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0 0] x4 + [1 1] x5 + [1]
                                  [0 0]      [0 0]      [0 0]      [0 0]      [0 0]      [1]
         e2(x1, x2, x3, x4, x5) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [1 1] x4 + [0 0] x5 + [0]
                                  [1 0]      [0 0]      [0 0]      [0 0]      [1 0]      [0]
         e3(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11) = [1 1] x1 + [0 0] x2 + [1 1] x3 + [0 0] x4 + [1 1] x5 + [0 0] x6 + [1 1] x7 + [0 0] x8 + [0 0] x9 + [0 0] x10 + [0 0] x11 + [0]
                                                            [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [0 0]       [0 0]       [0]
         e4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11) = [1 1] x1 + [0 0] x2 + [1 1] x3 + [0 0] x4 + [1 1] x5 + [0 0] x6 + [1 1] x7 + [0 0] x8 + [0 0] x9 + [0 0] x10 + [0 0] x11 + [0]
                                                            [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [0 0]       [0 0]       [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  g1() -> h1()
           , g1() -> h2()
           , g2() -> h1()
           , g2() -> h2()
           , e1(h1(), h2(), x, y, z) -> e2(x, x, y, z, z)
           , e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) ->
             e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)}
        Weak Trs:
          {  f2() -> g1()
           , f2() -> g2()
           , f1() -> g1()
           , f1() -> g2()}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component:
          {  g1() -> h1()
           , g1() -> h2()}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(e1) = {}, Uargs(e2) = {}, Uargs(e3) = {}, Uargs(e4) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           f1() = [2]
                  [0]
           g1() = [1]
                  [0]
           g2() = [0]
                  [0]
           f2() = [1]
                  [0]
           h1() = [0]
                  [0]
           h2() = [0]
                  [0]
           e1(x1, x2, x3, x4, x5) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0 0] x4 + [1 1] x5 + [1]
                                    [0 0]      [0 0]      [0 0]      [0 0]      [0 0]      [1]
           e2(x1, x2, x3, x4, x5) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0 1] x4 + [1 0] x5 + [0]
                                    [1 0]      [0 0]      [0 0]      [1 0]      [0 0]      [0]
           e3(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11) = [1 1] x1 + [0 0] x2 + [1 1] x3 + [0 0] x4 + [1 1] x5 + [0 0] x6 + [1 1] x7 + [0 0] x8 + [0 0] x9 + [0 0] x10 + [0 0] x11 + [0]
                                                              [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [0 0]       [0 0]       [0]
           e4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11) = [1 1] x1 + [0 0] x2 + [1 1] x3 + [0 0] x4 + [1 1] x5 + [0 0] x6 + [1 1] x7 + [0 0] x8 + [0 0] x9 + [0 0] x10 + [0 0] x11 + [0]
                                                              [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [0 0]       [0 0]       [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  g2() -> h1()
             , g2() -> h2()
             , e1(h1(), h2(), x, y, z) -> e2(x, x, y, z, z)
             , e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) ->
               e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)}
          Weak Trs:
            {  g1() -> h1()
             , g1() -> h2()
             , f2() -> g1()
             , f2() -> g2()
             , f1() -> g1()
             , f1() -> g2()}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The weightgap principle applies, where following rules are oriented strictly:
          
          TRS Component:
            {  g2() -> h1()
             , g2() -> h2()}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(e1) = {}, Uargs(e2) = {}, Uargs(e3) = {}, Uargs(e4) = {}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             f1() = [2]
                    [0]
             g1() = [0]
                    [0]
             g2() = [2]
                    [0]
             f2() = [2]
                    [0]
             h1() = [0]
                    [0]
             h2() = [0]
                    [0]
             e1(x1, x2, x3, x4, x5) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0 0] x4 + [1 1] x5 + [1]
                                      [0 0]      [0 0]      [0 0]      [0 0]      [0 0]      [1]
             e2(x1, x2, x3, x4, x5) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0 0] x4 + [1 1] x5 + [0]
                                      [1 0]      [0 0]      [0 0]      [1 0]      [0 0]      [0]
             e3(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11) = [1 1] x1 + [0 0] x2 + [1 1] x3 + [0 0] x4 + [1 1] x5 + [0 0] x6 + [1 1] x7 + [0 0] x8 + [0 0] x9 + [0 0] x10 + [0 0] x11 + [0]
                                                                [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [0 0]       [0 0]       [0]
             e4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11) = [1 1] x1 + [0 0] x2 + [1 1] x3 + [0 0] x4 + [1 1] x5 + [0 0] x6 + [1 0] x7 + [0 1] x8 + [0 0] x9 + [0 0] x10 + [0 0] x11 + [0]
                                                                [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [0 0]       [0 0]       [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Strict Trs:
              {  e1(h1(), h2(), x, y, z) -> e2(x, x, y, z, z)
               , e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) ->
                 e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)}
            Weak Trs:
              {  g2() -> h1()
               , g2() -> h2()
               , g1() -> h1()
               , g1() -> h2()
               , f2() -> g1()
               , f2() -> g2()
               , f1() -> g1()
               , f1() -> g2()}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            The weightgap principle applies, where following rules are oriented strictly:
            
            TRS Component:
              {e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) ->
               e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)}
            
            Interpretation of nonconstant growth:
            -------------------------------------
              The following argument positions are usable:
                Uargs(e1) = {}, Uargs(e2) = {}, Uargs(e3) = {}, Uargs(e4) = {}
              We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
              Interpretation Functions:
               f1() = [0]
                      [0]
               g1() = [0]
                      [0]
               g2() = [0]
                      [0]
               f2() = [0]
                      [0]
               h1() = [0]
                      [0]
               h2() = [0]
                      [0]
               e1(x1, x2, x3, x4, x5) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0 0] x4 + [1 1] x5 + [1]
                                        [0 0]      [0 0]      [0 0]      [0 0]      [0 0]      [1]
               e2(x1, x2, x3, x4, x5) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [1 1] x4 + [0 0] x5 + [0]
                                        [1 0]      [0 0]      [0 0]      [0 0]      [1 0]      [0]
               e3(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11) = [1 1] x1 + [0 0] x2 + [1 1] x3 + [0 0] x4 + [1 1] x5 + [0 0] x6 + [0 1] x7 + [1 0] x8 + [0 0] x9 + [0 0] x10 + [0 0] x11 + [2]
                                                                  [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [1 0]      [0 0]      [0 0]      [0 0]       [0 0]       [0]
               e4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11) = [1 1] x1 + [0 0] x2 + [1 1] x3 + [0 0] x4 + [1 1] x5 + [0 0] x6 + [1 1] x7 + [0 0] x8 + [0 0] x9 + [0 0] x10 + [0 0] x11 + [0]
                                                                  [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [1 0]      [0 0]      [0 0]       [0 0]       [0]
            
            The strictly oriented rules are moved into the weak component.
            
            We consider the following Problem:
            
              Strict Trs: {e1(h1(), h2(), x, y, z) -> e2(x, x, y, z, z)}
              Weak Trs:
                {  e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) ->
                   e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
                 , g2() -> h1()
                 , g2() -> h2()
                 , g1() -> h1()
                 , g1() -> h2()
                 , f2() -> g1()
                 , f2() -> g2()
                 , f1() -> g1()
                 , f1() -> g2()}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^1))
            
            Proof:
              The weightgap principle applies, where following rules are oriented strictly:
              
              TRS Component: {e1(h1(), h2(), x, y, z) -> e2(x, x, y, z, z)}
              
              Interpretation of nonconstant growth:
              -------------------------------------
                The following argument positions are usable:
                  Uargs(e1) = {}, Uargs(e2) = {}, Uargs(e3) = {}, Uargs(e4) = {}
                We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                Interpretation Functions:
                 f1() = [2]
                        [0]
                 g1() = [1]
                        [0]
                 g2() = [1]
                        [0]
                 f2() = [1]
                        [0]
                 h1() = [1]
                        [0]
                 h2() = [0]
                        [0]
                 e1(x1, x2, x3, x4, x5) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [0 0] x4 + [0 0] x5 + [0]
                                          [0 0]      [1 0]      [1 0]      [0 0]      [0 1]      [1]
                 e2(x1, x2, x3, x4, x5) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0 0] x4 + [0 0] x5 + [0]
                                          [1 0]      [0 0]      [0 0]      [0 1]      [0 0]      [0]
                 e3(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11) = [0 0] x1 + [1 0] x2 + [0 0] x3 + [1 0] x4 + [0 0] x5 + [1 0] x6 + [0 0] x7 + [1 0] x8 + [0 0] x9 + [0 0] x10 + [0 0] x11 + [0]
                                                                    [0 1]      [1 0]      [0 1]      [1 0]      [0 1]      [1 0]      [0 1]      [1 0]      [0 0]      [0 0]       [0 0]       [0]
                 e4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11) = [0 0] x1 + [1 0] x2 + [0 0] x3 + [1 0] x4 + [1 0] x5 + [0 0] x6 + [0 0] x7 + [1 0] x8 + [0 0] x9 + [0 0] x10 + [0 0] x11 + [0]
                                                                    [0 1]      [1 0]      [0 1]      [1 0]      [1 0]      [0 1]      [0 1]      [1 0]      [0 0]      [0 0]       [0 0]       [0]
              
              The strictly oriented rules are moved into the weak component.
              
              We consider the following Problem:
              
                Weak Trs:
                  {  e1(h1(), h2(), x, y, z) -> e2(x, x, y, z, z)
                   , e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) ->
                     e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
                   , g2() -> h1()
                   , g2() -> h2()
                   , g1() -> h1()
                   , g1() -> h2()
                   , f2() -> g1()
                   , f2() -> g2()
                   , f1() -> g1()
                   , f1() -> g2()}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(O(1),O(1))
              
              Proof:
                We consider the following Problem:
                
                  Weak Trs:
                    {  e1(h1(), h2(), x, y, z) -> e2(x, x, y, z, z)
                     , e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) ->
                       e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
                     , g2() -> h1()
                     , g2() -> h2()
                     , g1() -> h1()
                     , g1() -> h2()
                     , f2() -> g1()
                     , f2() -> g2()
                     , f1() -> g1()
                     , f1() -> g2()}
                  StartTerms: basic terms
                  Strategy: innermost
                
                Certificate: YES(O(1),O(1))
                
                Proof:
                  Empty rules are trivially bounded

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