YES

Problem:
 p(m,n,s(r)) -> p(m,r,n)
 p(m,s(n),0()) -> p(0(),n,m)
 p(m,0(),0()) -> m

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
                     [1 0 0]     [1 0 0]     [1 0 0]  
   [p](x0, x1, x2) = [0 1 1]x0 + [0 1 1]x1 + [0 1 1]x2
                     [0 1 1]     [0 1 1]     [0 1 1]  ,
   
             [1 0 0]     [1]
   [s](x0) = [0 0 1]x0 + [0]
             [0 1 0]     [0],
   
         [0]
   [0] = [0]
         [0]
  orientation:
                 [1 0 0]    [1 0 0]    [1 0 0]    [1]    [1 0 0]    [1 0 0]    [1 0 0]            
   p(m,n,s(r)) = [0 1 1]m + [0 1 1]n + [0 1 1]r + [0] >= [0 1 1]m + [0 1 1]n + [0 1 1]r = p(m,r,n)
                 [0 1 1]    [0 1 1]    [0 1 1]    [0]    [0 1 1]    [0 1 1]    [0 1 1]            
   
                   [1 0 0]    [1 0 0]    [1]    [1 0 0]    [1 0 0]              
   p(m,s(n),0()) = [0 1 1]m + [0 1 1]n + [0] >= [0 1 1]m + [0 1 1]n = p(0(),n,m)
                   [0 1 1]    [0 1 1]    [0]    [0 1 1]    [0 1 1]              
   
                  [1 0 0]          
   p(m,0(),0()) = [0 1 1]m >= m = m
                  [0 1 1]          
  problem:
   p(m,0(),0()) -> m
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                           [1 0 0]     [1 1 0]  
    [p](x0, x1, x2) = x0 + [0 0 0]x1 + [0 0 0]x2
                           [0 0 0]     [0 0 0]  ,
    
          [0]
    [0] = [1]
          [0]
   orientation:
                       [1]         
    p(m,0(),0()) = m + [0] >= m = m
                       [0]         
   problem:
    
   Qed