YES

Problem:
 p(0()) -> 0()
 p(s(x)) -> x
 minus(x,0()) -> x
 minus(x,s(y)) -> minus(p(x),y)

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