We consider the following Problem:

  Strict Trs:
    {  gcd(x, 0()) -> x
     , gcd(0(), y) -> y
     , gcd(s(x), s(y)) ->
       if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y)))}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  gcd(x, 0()) -> x
       , gcd(0(), y) -> y
       , gcd(s(x), s(y)) ->
         if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y)))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {gcd(s(x), s(y)) ->
       if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y)))}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(gcd) = {}, Uargs(s) = {}, Uargs(if) = {}, Uargs(<) = {},
        Uargs(-) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       gcd(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                     [0 0]      [0 0]      [1]
       0() = [0]
             [0]
       s(x1) = [0 0] x1 + [0]
               [0 0]      [0]
       if(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
                        [0 0]      [0 0]      [0 0]      [1]
       <(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                      [0 0]      [0 0]      [0]
       -(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                   [0 0]      [0 0]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  gcd(x, 0()) -> x
         , gcd(0(), y) -> y}
      Weak Trs:
        {gcd(s(x), s(y)) ->
         if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y)))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {gcd(0(), y) -> y}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(gcd) = {}, Uargs(s) = {}, Uargs(if) = {}, Uargs(<) = {},
          Uargs(-) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         gcd(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                       [0 0]      [0 1]      [1]
         0() = [0]
               [0]
         s(x1) = [0 0] x1 + [0]
                 [0 0]      [0]
         if(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [1]
                          [0 0]      [0 0]      [0 1]      [0]
         <(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                        [0 0]      [0 0]      [0]
         -(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                     [0 0]      [0 0]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs: {gcd(x, 0()) -> x}
        Weak Trs:
          {  gcd(0(), y) -> y
           , gcd(s(x), s(y)) ->
             if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y)))}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {gcd(x, 0()) -> x}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(gcd) = {}, Uargs(s) = {}, Uargs(if) = {}, Uargs(<) = {},
            Uargs(-) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           gcd(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                         [0 1]      [0 1]      [1]
           0() = [0]
                 [0]
           s(x1) = [0 0] x1 + [0]
                   [0 0]      [0]
           if(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [1]
                            [0 0]      [0 0]      [0 0]      [1]
           <(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                          [0 0]      [0 0]      [0]
           -(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                       [0 0]      [0 0]      [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Weak Trs:
            {  gcd(x, 0()) -> x
             , gcd(0(), y) -> y
             , gcd(s(x), s(y)) ->
               if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y)))}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(O(1),O(1))
        
        Proof:
          We consider the following Problem:
          
            Weak Trs:
              {  gcd(x, 0()) -> x
               , gcd(0(), y) -> y
               , gcd(s(x), s(y)) ->
                 if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y)))}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(O(1),O(1))
          
          Proof:
            Empty rules are trivially bounded

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