We consider the following Problem:

  Strict Trs:
    {  minus(0()) -> 0()
     , +(x, 0()) -> x
     , +(0(), y) -> y
     , +(minus(1()), 1()) -> 0()
     , minus(minus(x)) -> x
     , +(x, minus(y)) -> minus(+(minus(x), y))
     , +(x, +(y, z)) -> +(+(x, y), z)
     , +(minus(+(x, 1())), 1()) -> minus(x)}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  minus(0()) -> 0()
       , +(x, 0()) -> x
       , +(0(), y) -> y
       , +(minus(1()), 1()) -> 0()
       , minus(minus(x)) -> x
       , +(x, minus(y)) -> minus(+(minus(x), y))
       , +(x, +(y, z)) -> +(+(x, y), z)
       , +(minus(+(x, 1())), 1()) -> minus(x)}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  minus(0()) -> 0()
       , +(minus(1()), 1()) -> 0()
       , minus(minus(x)) -> x}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(minus) = {1}, Uargs(+) = {1}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       minus(x1) = [1 0] x1 + [1]
                   [0 1]      [1]
       0() = [0]
             [0]
       +(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                   [0 0]      [0 0]      [1]
       1() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  +(x, 0()) -> x
         , +(0(), y) -> y
         , +(x, minus(y)) -> minus(+(minus(x), y))
         , +(x, +(y, z)) -> +(+(x, y), z)
         , +(minus(+(x, 1())), 1()) -> minus(x)}
      Weak Trs:
        {  minus(0()) -> 0()
         , +(minus(1()), 1()) -> 0()
         , minus(minus(x)) -> x}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {+(0(), y) -> y}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(minus) = {1}, Uargs(+) = {1}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         minus(x1) = [1 0] x1 + [0]
                     [0 1]      [2]
         0() = [0]
               [0]
         +(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                     [1 0]      [0 1]      [0]
         1() = [0]
               [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  +(x, 0()) -> x
           , +(x, minus(y)) -> minus(+(minus(x), y))
           , +(x, +(y, z)) -> +(+(x, y), z)
           , +(minus(+(x, 1())), 1()) -> minus(x)}
        Weak Trs:
          {  +(0(), y) -> y
           , minus(0()) -> 0()
           , +(minus(1()), 1()) -> 0()
           , minus(minus(x)) -> x}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component:
          {  +(x, 0()) -> x
           , +(minus(+(x, 1())), 1()) -> minus(x)}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(minus) = {1}, Uargs(+) = {1}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           minus(x1) = [1 0] x1 + [0]
                       [0 1]      [0]
           0() = [0]
                 [0]
           +(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                       [0 1]      [0 1]      [1]
           1() = [0]
                 [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  +(x, minus(y)) -> minus(+(minus(x), y))
             , +(x, +(y, z)) -> +(+(x, y), z)}
          Weak Trs:
            {  +(x, 0()) -> x
             , +(minus(+(x, 1())), 1()) -> minus(x)
             , +(0(), y) -> y
             , minus(0()) -> 0()
             , +(minus(1()), 1()) -> 0()
             , minus(minus(x)) -> x}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          We consider the following Problem:
          
            Strict Trs:
              {  +(x, minus(y)) -> minus(+(minus(x), y))
               , +(x, +(y, z)) -> +(+(x, y), z)}
            Weak Trs:
              {  +(x, 0()) -> x
               , +(minus(+(x, 1())), 1()) -> minus(x)
               , +(0(), y) -> y
               , minus(0()) -> 0()
               , +(minus(1()), 1()) -> 0()
               , minus(minus(x)) -> 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:
            {  minus_0(2) -> 1
             , 0_0() -> 1
             , 0_0() -> 2
             , +_0(2, 2) -> 1
             , 1_0() -> 1
             , 1_0() -> 2}

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