We consider the following Problem:

  Strict Trs:
    {  active(U11(tt(), M, N)) -> mark(U12(tt(), M, N))
     , active(U12(tt(), M, N)) -> mark(s(plus(N, M)))
     , active(plus(N, 0())) -> mark(N)
     , active(plus(N, s(M))) -> mark(U11(tt(), M, N))
     , mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3))
     , mark(tt()) -> active(tt())
     , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3))
     , mark(s(X)) -> active(s(mark(X)))
     , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
     , mark(0()) -> active(0())
     , U11(mark(X1), X2, X3) -> U11(X1, X2, X3)
     , U11(X1, mark(X2), X3) -> U11(X1, X2, X3)
     , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3)
     , U11(active(X1), X2, X3) -> U11(X1, X2, X3)
     , U11(X1, active(X2), X3) -> U11(X1, X2, X3)
     , U11(X1, X2, active(X3)) -> U11(X1, X2, X3)
     , U12(mark(X1), X2, X3) -> U12(X1, X2, X3)
     , U12(X1, mark(X2), X3) -> U12(X1, X2, X3)
     , U12(X1, X2, mark(X3)) -> U12(X1, X2, X3)
     , U12(active(X1), X2, X3) -> U12(X1, X2, X3)
     , U12(X1, active(X2), X3) -> U12(X1, X2, X3)
     , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)
     , s(mark(X)) -> s(X)
     , s(active(X)) -> s(X)
     , plus(mark(X1), X2) -> plus(X1, X2)
     , plus(X1, mark(X2)) -> plus(X1, X2)
     , plus(active(X1), X2) -> plus(X1, X2)
     , plus(X1, active(X2)) -> plus(X1, X2)}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  active(U11(tt(), M, N)) -> mark(U12(tt(), M, N))
       , active(U12(tt(), M, N)) -> mark(s(plus(N, M)))
       , active(plus(N, 0())) -> mark(N)
       , active(plus(N, s(M))) -> mark(U11(tt(), M, N))
       , mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3))
       , mark(tt()) -> active(tt())
       , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3))
       , mark(s(X)) -> active(s(mark(X)))
       , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
       , mark(0()) -> active(0())
       , U11(mark(X1), X2, X3) -> U11(X1, X2, X3)
       , U11(X1, mark(X2), X3) -> U11(X1, X2, X3)
       , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3)
       , U11(active(X1), X2, X3) -> U11(X1, X2, X3)
       , U11(X1, active(X2), X3) -> U11(X1, X2, X3)
       , U11(X1, X2, active(X3)) -> U11(X1, X2, X3)
       , U12(mark(X1), X2, X3) -> U12(X1, X2, X3)
       , U12(X1, mark(X2), X3) -> U12(X1, X2, X3)
       , U12(X1, X2, mark(X3)) -> U12(X1, X2, X3)
       , U12(active(X1), X2, X3) -> U12(X1, X2, X3)
       , U12(X1, active(X2), X3) -> U12(X1, X2, X3)
       , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)
       , s(mark(X)) -> s(X)
       , s(active(X)) -> s(X)
       , plus(mark(X1), X2) -> plus(X1, X2)
       , plus(X1, mark(X2)) -> plus(X1, X2)
       , plus(active(X1), X2) -> plus(X1, X2)
       , plus(X1, active(X2)) -> plus(X1, X2)}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  U12(X1, X2, mark(X3)) -> U12(X1, X2, X3)
       , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(active) = {1}, Uargs(U11) = {1}, Uargs(mark) = {1},
        Uargs(U12) = {1}, Uargs(s) = {1}, Uargs(plus) = {1, 2}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       active(x1) = [1 1] x1 + [0]
                    [0 0]      [1]
       U11(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [1 0] x3 + [0]
                         [0 1]      [1 1]      [0 1]      [0]
       tt() = [0]
              [0]
       mark(x1) = [1 1] x1 + [1]
                  [0 0]      [0]
       U12(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [1 1] x3 + [0]
                         [0 1]      [1 1]      [0 0]      [0]
       s(x1) = [1 0] x1 + [0]
               [0 1]      [0]
       plus(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                      [0 1]      [0 1]      [0]
       0() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  active(U11(tt(), M, N)) -> mark(U12(tt(), M, N))
         , active(U12(tt(), M, N)) -> mark(s(plus(N, M)))
         , active(plus(N, 0())) -> mark(N)
         , active(plus(N, s(M))) -> mark(U11(tt(), M, N))
         , mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3))
         , mark(tt()) -> active(tt())
         , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3))
         , mark(s(X)) -> active(s(mark(X)))
         , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
         , mark(0()) -> active(0())
         , U11(mark(X1), X2, X3) -> U11(X1, X2, X3)
         , U11(X1, mark(X2), X3) -> U11(X1, X2, X3)
         , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3)
         , U11(active(X1), X2, X3) -> U11(X1, X2, X3)
         , U11(X1, active(X2), X3) -> U11(X1, X2, X3)
         , U11(X1, X2, active(X3)) -> U11(X1, X2, X3)
         , U12(mark(X1), X2, X3) -> U12(X1, X2, X3)
         , U12(X1, mark(X2), X3) -> U12(X1, X2, X3)
         , U12(active(X1), X2, X3) -> U12(X1, X2, X3)
         , U12(X1, active(X2), X3) -> U12(X1, X2, X3)
         , s(mark(X)) -> s(X)
         , s(active(X)) -> s(X)
         , plus(mark(X1), X2) -> plus(X1, X2)
         , plus(X1, mark(X2)) -> plus(X1, X2)
         , plus(active(X1), X2) -> plus(X1, X2)
         , plus(X1, active(X2)) -> plus(X1, X2)}
      Weak Trs:
        {  U12(X1, X2, mark(X3)) -> U12(X1, X2, X3)
         , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component:
        {  U12(X1, mark(X2), X3) -> U12(X1, X2, X3)
         , U12(X1, active(X2), X3) -> U12(X1, X2, X3)
         , plus(mark(X1), X2) -> plus(X1, X2)
         , plus(active(X1), X2) -> plus(X1, X2)}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(active) = {1}, Uargs(U11) = {1}, Uargs(mark) = {1},
          Uargs(U12) = {1}, Uargs(s) = {1}, Uargs(plus) = {1, 2}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         active(x1) = [1 1] x1 + [1]
                      [0 0]      [3]
         U11(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [1 0] x3 + [0]
                           [0 1]      [1 1]      [0 1]      [0]
         tt() = [0]
                [0]
         mark(x1) = [1 1] x1 + [1]
                    [0 0]      [0]
         U12(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [0 0] x3 + [0]
                           [0 1]      [0 0]      [1 1]      [0]
         s(x1) = [1 0] x1 + [0]
                 [0 1]      [0]
         plus(x1, x2) = [1 1] x1 + [1 0] x2 + [0]
                        [0 0]      [0 1]      [0]
         0() = [0]
               [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  active(U11(tt(), M, N)) -> mark(U12(tt(), M, N))
           , active(U12(tt(), M, N)) -> mark(s(plus(N, M)))
           , active(plus(N, 0())) -> mark(N)
           , active(plus(N, s(M))) -> mark(U11(tt(), M, N))
           , mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3))
           , mark(tt()) -> active(tt())
           , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3))
           , mark(s(X)) -> active(s(mark(X)))
           , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
           , mark(0()) -> active(0())
           , U11(mark(X1), X2, X3) -> U11(X1, X2, X3)
           , U11(X1, mark(X2), X3) -> U11(X1, X2, X3)
           , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3)
           , U11(active(X1), X2, X3) -> U11(X1, X2, X3)
           , U11(X1, active(X2), X3) -> U11(X1, X2, X3)
           , U11(X1, X2, active(X3)) -> U11(X1, X2, X3)
           , U12(mark(X1), X2, X3) -> U12(X1, X2, X3)
           , U12(active(X1), X2, X3) -> U12(X1, X2, X3)
           , s(mark(X)) -> s(X)
           , s(active(X)) -> s(X)
           , plus(X1, mark(X2)) -> plus(X1, X2)
           , plus(X1, active(X2)) -> plus(X1, X2)}
        Weak Trs:
          {  U12(X1, mark(X2), X3) -> U12(X1, X2, X3)
           , U12(X1, active(X2), X3) -> U12(X1, X2, X3)
           , plus(mark(X1), X2) -> plus(X1, X2)
           , plus(active(X1), X2) -> plus(X1, X2)
           , U12(X1, X2, mark(X3)) -> U12(X1, X2, X3)
           , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component:
          {  active(U11(tt(), M, N)) -> mark(U12(tt(), M, N))
           , active(U12(tt(), M, N)) -> mark(s(plus(N, M)))
           , active(plus(N, 0())) -> mark(N)
           , active(plus(N, s(M))) -> mark(U11(tt(), M, N))}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(active) = {1}, Uargs(U11) = {1}, Uargs(mark) = {1},
            Uargs(U12) = {1}, Uargs(s) = {1}, Uargs(plus) = {1, 2}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           active(x1) = [1 1] x1 + [1]
                        [0 0]      [1]
           U11(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [1 0] x3 + [0]
                             [0 1]      [1 1]      [0 1]      [0]
           tt() = [0]
                  [0]
           mark(x1) = [1 1] x1 + [0]
                      [0 0]      [1]
           U12(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [1 1] x3 + [0]
                             [0 1]      [1 1]      [0 0]      [0]
           s(x1) = [1 0] x1 + [0]
                   [0 1]      [0]
           plus(x1, x2) = [1 1] x1 + [1 0] x2 + [0]
                          [0 0]      [0 1]      [0]
           0() = [0]
                 [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3))
             , mark(tt()) -> active(tt())
             , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3))
             , mark(s(X)) -> active(s(mark(X)))
             , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
             , mark(0()) -> active(0())
             , U11(mark(X1), X2, X3) -> U11(X1, X2, X3)
             , U11(X1, mark(X2), X3) -> U11(X1, X2, X3)
             , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3)
             , U11(active(X1), X2, X3) -> U11(X1, X2, X3)
             , U11(X1, active(X2), X3) -> U11(X1, X2, X3)
             , U11(X1, X2, active(X3)) -> U11(X1, X2, X3)
             , U12(mark(X1), X2, X3) -> U12(X1, X2, X3)
             , U12(active(X1), X2, X3) -> U12(X1, X2, X3)
             , s(mark(X)) -> s(X)
             , s(active(X)) -> s(X)
             , plus(X1, mark(X2)) -> plus(X1, X2)
             , plus(X1, active(X2)) -> plus(X1, X2)}
          Weak Trs:
            {  active(U11(tt(), M, N)) -> mark(U12(tt(), M, N))
             , active(U12(tt(), M, N)) -> mark(s(plus(N, M)))
             , active(plus(N, 0())) -> mark(N)
             , active(plus(N, s(M))) -> mark(U11(tt(), M, N))
             , U12(X1, mark(X2), X3) -> U12(X1, X2, X3)
             , U12(X1, active(X2), X3) -> U12(X1, X2, X3)
             , plus(mark(X1), X2) -> plus(X1, X2)
             , plus(active(X1), X2) -> plus(X1, X2)
             , U12(X1, X2, mark(X3)) -> U12(X1, X2, X3)
             , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The weightgap principle applies, where following rules are oriented strictly:
          
          TRS Component:
            {  U11(mark(X1), X2, X3) -> U11(X1, X2, X3)
             , U11(X1, mark(X2), X3) -> U11(X1, X2, X3)
             , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3)
             , U11(active(X1), X2, X3) -> U11(X1, X2, X3)
             , U11(X1, active(X2), X3) -> U11(X1, X2, X3)
             , U11(X1, X2, active(X3)) -> U11(X1, X2, X3)
             , U12(mark(X1), X2, X3) -> U12(X1, X2, X3)
             , U12(active(X1), X2, X3) -> U12(X1, X2, X3)
             , s(mark(X)) -> s(X)
             , s(active(X)) -> s(X)
             , plus(X1, mark(X2)) -> plus(X1, X2)
             , plus(X1, active(X2)) -> plus(X1, X2)}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(active) = {1}, Uargs(U11) = {1}, Uargs(mark) = {1},
              Uargs(U12) = {1}, Uargs(s) = {1}, Uargs(plus) = {1, 2}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             active(x1) = [1 0] x1 + [2]
                          [1 0]      [3]
             U11(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [0]
                               [0 0]      [0 0]      [0 0]      [1]
             tt() = [2]
                    [0]
             mark(x1) = [1 0] x1 + [1]
                        [1 0]      [1]
             U12(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
                               [0 0]      [0 0]      [0 0]      [1]
             s(x1) = [1 0] x1 + [0]
                     [0 0]      [1]
             plus(x1, x2) = [1 0] x1 + [1 0] x2 + [2]
                            [1 0]      [0 0]      [1]
             0() = [0]
                   [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Strict Trs:
              {  mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3))
               , mark(tt()) -> active(tt())
               , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3))
               , mark(s(X)) -> active(s(mark(X)))
               , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
               , mark(0()) -> active(0())}
            Weak Trs:
              {  U11(mark(X1), X2, X3) -> U11(X1, X2, X3)
               , U11(X1, mark(X2), X3) -> U11(X1, X2, X3)
               , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3)
               , U11(active(X1), X2, X3) -> U11(X1, X2, X3)
               , U11(X1, active(X2), X3) -> U11(X1, X2, X3)
               , U11(X1, X2, active(X3)) -> U11(X1, X2, X3)
               , U12(mark(X1), X2, X3) -> U12(X1, X2, X3)
               , U12(active(X1), X2, X3) -> U12(X1, X2, X3)
               , s(mark(X)) -> s(X)
               , s(active(X)) -> s(X)
               , plus(X1, mark(X2)) -> plus(X1, X2)
               , plus(X1, active(X2)) -> plus(X1, X2)
               , active(U11(tt(), M, N)) -> mark(U12(tt(), M, N))
               , active(U12(tt(), M, N)) -> mark(s(plus(N, M)))
               , active(plus(N, 0())) -> mark(N)
               , active(plus(N, s(M))) -> mark(U11(tt(), M, N))
               , U12(X1, mark(X2), X3) -> U12(X1, X2, X3)
               , U12(X1, active(X2), X3) -> U12(X1, X2, X3)
               , plus(mark(X1), X2) -> plus(X1, X2)
               , plus(active(X1), X2) -> plus(X1, X2)
               , U12(X1, X2, mark(X3)) -> U12(X1, X2, X3)
               , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            The weightgap principle applies, where following rules are oriented strictly:
            
            TRS Component:
              {  mark(tt()) -> active(tt())
               , mark(0()) -> active(0())}
            
            Interpretation of nonconstant growth:
            -------------------------------------
              The following argument positions are usable:
                Uargs(active) = {1}, Uargs(U11) = {1}, Uargs(mark) = {1},
                Uargs(U12) = {1}, Uargs(s) = {1}, Uargs(plus) = {1, 2}
              We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
              Interpretation Functions:
               active(x1) = [1 2] x1 + [0]
                            [0 0]      [0]
               U11(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [2]
                                 [0 0]      [0 0]      [0 0]      [1]
               tt() = [0]
                      [0]
               mark(x1) = [1 0] x1 + [1]
                          [0 0]      [0]
               U12(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [2]
                                 [0 0]      [0 0]      [0 0]      [1]
               s(x1) = [1 0] x1 + [0]
                       [0 0]      [0]
               plus(x1, x2) = [1 0] x1 + [1 0] x2 + [3]
                              [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:
                {  mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3))
                 , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3))
                 , mark(s(X)) -> active(s(mark(X)))
                 , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))}
              Weak Trs:
                {  mark(tt()) -> active(tt())
                 , mark(0()) -> active(0())
                 , U11(mark(X1), X2, X3) -> U11(X1, X2, X3)
                 , U11(X1, mark(X2), X3) -> U11(X1, X2, X3)
                 , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3)
                 , U11(active(X1), X2, X3) -> U11(X1, X2, X3)
                 , U11(X1, active(X2), X3) -> U11(X1, X2, X3)
                 , U11(X1, X2, active(X3)) -> U11(X1, X2, X3)
                 , U12(mark(X1), X2, X3) -> U12(X1, X2, X3)
                 , U12(active(X1), X2, X3) -> U12(X1, X2, X3)
                 , s(mark(X)) -> s(X)
                 , s(active(X)) -> s(X)
                 , plus(X1, mark(X2)) -> plus(X1, X2)
                 , plus(X1, active(X2)) -> plus(X1, X2)
                 , active(U11(tt(), M, N)) -> mark(U12(tt(), M, N))
                 , active(U12(tt(), M, N)) -> mark(s(plus(N, M)))
                 , active(plus(N, 0())) -> mark(N)
                 , active(plus(N, s(M))) -> mark(U11(tt(), M, N))
                 , U12(X1, mark(X2), X3) -> U12(X1, X2, X3)
                 , U12(X1, active(X2), X3) -> U12(X1, X2, X3)
                 , plus(mark(X1), X2) -> plus(X1, X2)
                 , plus(active(X1), X2) -> plus(X1, X2)
                 , U12(X1, X2, mark(X3)) -> U12(X1, X2, X3)
                 , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^1))
            
            Proof:
              We consider the following Problem:
              
                Strict Trs:
                  {  mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3))
                   , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3))
                   , mark(s(X)) -> active(s(mark(X)))
                   , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))}
                Weak Trs:
                  {  mark(tt()) -> active(tt())
                   , mark(0()) -> active(0())
                   , U11(mark(X1), X2, X3) -> U11(X1, X2, X3)
                   , U11(X1, mark(X2), X3) -> U11(X1, X2, X3)
                   , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3)
                   , U11(active(X1), X2, X3) -> U11(X1, X2, X3)
                   , U11(X1, active(X2), X3) -> U11(X1, X2, X3)
                   , U11(X1, X2, active(X3)) -> U11(X1, X2, X3)
                   , U12(mark(X1), X2, X3) -> U12(X1, X2, X3)
                   , U12(active(X1), X2, X3) -> U12(X1, X2, X3)
                   , s(mark(X)) -> s(X)
                   , s(active(X)) -> s(X)
                   , plus(X1, mark(X2)) -> plus(X1, X2)
                   , plus(X1, active(X2)) -> plus(X1, X2)
                   , active(U11(tt(), M, N)) -> mark(U12(tt(), M, N))
                   , active(U12(tt(), M, N)) -> mark(s(plus(N, M)))
                   , active(plus(N, 0())) -> mark(N)
                   , active(plus(N, s(M))) -> mark(U11(tt(), M, N))
                   , U12(X1, mark(X2), X3) -> U12(X1, X2, X3)
                   , U12(X1, active(X2), X3) -> U12(X1, X2, X3)
                   , plus(mark(X1), X2) -> plus(X1, X2)
                   , plus(active(X1), X2) -> plus(X1, X2)
                   , U12(X1, X2, mark(X3)) -> U12(X1, X2, X3)
                   , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)}
                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:
                {  active_0(2) -> 1
                 , U11_0(2, 2, 2) -> 1
                 , tt_0() -> 2
                 , mark_0(2) -> 1
                 , U12_0(2, 2, 2) -> 1
                 , s_0(2) -> 1
                 , plus_0(2, 2) -> 1
                 , 0_0() -> 2}

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