YES

Problem:
 half(0()) -> 0()
 half(s(s(x))) -> s(half(x))
 log(s(0())) -> 0()
 log(s(s(x))) -> s(log(s(half(x))))

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
               [1 0 0]     [1]
   [log](x0) = [0 0 1]x0 + [0]
               [1 0 0]     [1],
   
             [1 0 0]  
   [s](x0) = [0 0 0]x0
             [1 0 0]  ,
   
                [1 0 0]     [0]
   [half](x0) = [0 0 0]x0 + [1]
                [1 0 0]     [1],
   
         [0]
   [0] = [0]
         [0]
  orientation:
               [0]    [0]      
   half(0()) = [1] >= [0] = 0()
               [1]    [0]      
   
                   [1 0 0]    [0]    [1 0 0]              
   half(s(s(x))) = [0 0 0]x + [1] >= [0 0 0]x = s(half(x))
                   [1 0 0]    [1]    [1 0 0]              
   
                 [1]    [0]      
   log(s(0())) = [0] >= [0] = 0()
                 [1]    [0]      
   
                  [1 0 0]    [1]    [1 0 0]    [1]                     
   log(s(s(x))) = [1 0 0]x + [0] >= [0 0 0]x + [0] = s(log(s(half(x))))
                  [1 0 0]    [1]    [1 0 0]    [1]                     
  problem:
   half(0()) -> 0()
   half(s(s(x))) -> s(half(x))
   log(s(s(x))) -> s(log(s(half(x))))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [log](x0) = x0,
    
    [s](x0) = 2x0 + 6,
    
    [half](x0) = x0,
    
    [0] = 1
   orientation:
    half(0()) = 1 >= 1 = 0()
    
    half(s(s(x))) = 4x + 18 >= 2x + 6 = s(half(x))
    
    log(s(s(x))) = 4x + 18 >= 4x + 18 = s(log(s(half(x))))
   problem:
    half(0()) -> 0()
    log(s(s(x))) -> s(log(s(half(x))))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                 [1 0 1]  
     [log](x0) = [1 0 1]x0
                 [1 0 1]  ,
     
               [1 0 0]     [0]
     [s](x0) = [0 0 1]x0 + [1]
               [0 1 0]     [0],
     
                  [1 0 1]  
     [half](x0) = [0 0 0]x0
                  [0 1 1]  ,
     
           [0]
     [0] = [0]
           [1]
    orientation:
                 [1]    [0]      
     half(0()) = [0] >= [0] = 0()
                 [1]    [1]      
     
                    [1 0 1]    [1]    [1 0 1]    [0]                     
     log(s(s(x))) = [1 0 1]x + [1] >= [1 0 1]x + [1] = s(log(s(half(x))))
                    [1 0 1]    [1]    [1 0 1]    [0]                     
    problem:
     
    Qed