YES

Problem:
 sum(0()) -> 0()
 sum(s(x)) -> +(sum(x),s(x))
 sum1(0()) -> 0()
 sum1(s(x)) -> s(+(sum1(x),+(x,x)))

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
                [1 1 1]  
   [sum1](x0) = [1 1 1]x0
                [1 1 1]  ,
   
                 [1 0 0]     [1 0 0]  
   [+](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                 [0 0 0]     [0 1 1]  ,
   
             [1 1 0]  
   [s](x0) = [1 1 1]x0
             [1 0 1]  ,
   
               [1 0 1]     [1]
   [sum](x0) = [0 0 0]x0 + [0]
               [1 1 1]     [0],
   
         [0]
   [0] = [0]
         [0]
  orientation:
              [1]    [0]      
   sum(0()) = [0] >= [0] = 0()
              [0]    [0]      
   
               [2 1 1]    [1]    [2 1 1]    [1]                 
   sum(s(x)) = [0 0 0]x + [0] >= [0 0 0]x + [0] = +(sum(x),s(x))
               [3 2 2]    [0]    [2 1 2]    [0]                 
   
               [0]    [0]      
   sum1(0()) = [0] >= [0] = 0()
               [0]    [0]      
   
                [3 2 2]     [3 1 1]                        
   sum1(s(x)) = [3 2 2]x >= [3 2 2]x = s(+(sum1(x),+(x,x)))
                [3 2 2]     [3 2 2]                        
  problem:
   sum(s(x)) -> +(sum(x),s(x))
   sum1(0()) -> 0()
   sum1(s(x)) -> s(+(sum1(x),+(x,x)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                 [1 1 1]     [0]
    [sum1](x0) = [1 1 1]x0 + [1]
                 [1 1 1]     [1],
    
                  [1 0 0]     [1 0 0]  
    [+](x0, x1) = [0 0 0]x0 + [0 1 1]x1
                  [0 0 0]     [0 0 0]  ,
    
              [1 1 0]     [0]
    [s](x0) = [1 0 1]x0 + [1]
              [1 1 1]     [1],
    
                [1 1 1]  
    [sum](x0) = [0 1 1]x0
                [1 1 1]  ,
    
          [1]
    [0] = [0]
          [0]
   orientation:
                [3 2 2]    [2]    [2 2 1]    [0]                 
    sum(s(x)) = [2 1 2]x + [2] >= [2 1 2]x + [2] = +(sum(x),s(x))
                [3 2 2]    [2]    [0 0 0]    [0]                 
    
                [1]    [1]      
    sum1(0()) = [2] >= [0] = 0()
                [2]    [0]      
    
                 [3 2 2]    [2]    [3 2 2]    [0]                       
    sum1(s(x)) = [3 2 2]x + [3] >= [3 1 1]x + [1] = s(+(sum1(x),+(x,x)))
                 [3 2 2]    [3]    [3 2 2]    [1]                       
   problem:
    sum1(0()) -> 0()
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                  [1 0 0]     [1]
     [sum1](x0) = [0 0 0]x0 + [0]
                  [0 0 0]     [0],
     
           [0]
     [0] = [0]
           [0]
    orientation:
                 [1]    [0]      
     sum1(0()) = [0] >= [0] = 0()
                 [0]    [0]      
    problem:
     
    Qed