YES

Problem:
 double(0()) -> 0()
 double(s(x)) -> s(s(double(x)))
 half(0()) -> 0()
 half(s(0())) -> 0()
 half(s(s(x))) -> s(half(x))
 -(x,0()) -> x
 -(s(x),s(y)) -> -(x,y)
 if(0(),y,z) -> y
 if(s(x),y,z) -> z
 half(double(x)) -> x

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [-](x0, x1) = 4x0 + 2x1 + 6,
   
   [double](x0) = 2x0,
   
   [half](x0) = x0 + 5,
   
   [if](x0, x1, x2) = 2x0 + 4x1 + 4x2 + 1,
   
   [0] = 0,
   
   [s](x0) = x0 + 1
  orientation:
   double(0()) = 0 >= 0 = 0()
   
   double(s(x)) = 2x + 2 >= 2x + 2 = s(s(double(x)))
   
   half(0()) = 5 >= 0 = 0()
   
   half(s(0())) = 6 >= 0 = 0()
   
   half(s(s(x))) = x + 7 >= x + 6 = s(half(x))
   
   -(x,0()) = 4x + 6 >= x = x
   
   -(s(x),s(y)) = 4x + 2y + 12 >= 4x + 2y + 6 = -(x,y)
   
   if(0(),y,z) = 4y + 4z + 1 >= y = y
   
   if(s(x),y,z) = 2x + 4y + 4z + 3 >= z = z
   
   half(double(x)) = 2x + 5 >= x = x
  problem:
   double(0()) -> 0()
   double(s(x)) -> s(s(double(x)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                   [1 0 1]  
    [double](x0) = [0 1 0]x0
                   [0 0 1]  ,
    
          [0]
    [0] = [1]
          [1],
    
              [1 0 0]  
    [s](x0) = [0 0 0]x0
              [0 0 1]  
   orientation:
                  [1]    [0]      
    double(0()) = [1] >= [1] = 0()
                  [1]    [1]      
    
                   [1 0 1]     [1 0 1]                   
    double(s(x)) = [0 0 0]x >= [0 0 0]x = s(s(double(x)))
                   [0 0 1]     [0 0 1]                   
   problem:
    double(s(x)) -> s(s(double(x)))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                    [1 1 1]  
     [double](x0) = [0 1 1]x0
                    [0 1 1]  ,
     
               [1 0 0]     [0]
     [s](x0) = [0 0 1]x0 + [0]
               [0 1 0]     [1]
    orientation:
                    [1 1 1]    [1]    [1 1 1]    [0]                  
     double(s(x)) = [0 1 1]x + [1] >= [0 1 1]x + [1] = s(s(double(x)))
                    [0 1 1]    [1]    [0 1 1]    [1]                  
    problem:
     
    Qed