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:
 DP Processor:
  DPs:
   p#(m,n,s(r)) -> p#(m,r,n)
   p#(m,s(n),0()) -> p#(0(),n,m)
  TRS:
   p(m,n,s(r)) -> p(m,r,n)
   p(m,s(n),0()) -> p(0(),n,m)
   p(m,0(),0()) -> m
  Matrix Interpretation Processor:
   dimension: 1
   interpretation:
    [p#](x0, x1, x2) = x0 + x1 + x2 + 1,
    
    [0] = 0,
    
    [p](x0, x1, x2) = x0,
    
    [s](x0) = x0 + 1
   orientation:
    p#(m,n,s(r)) = m + n + r + 2 >= m + n + r + 1 = p#(m,r,n)
    
    p#(m,s(n),0()) = m + n + 2 >= m + n + 1 = p#(0(),n,m)
    
    p(m,n,s(r)) = m >= m = p(m,r,n)
    
    p(m,s(n),0()) = m >= 0 = p(0(),n,m)
    
    p(m,0(),0()) = m >= m = m
   problem:
    DPs:
     
    TRS:
     p(m,n,s(r)) -> p(m,r,n)
     p(m,s(n),0()) -> p(0(),n,m)
     p(m,0(),0()) -> m
   Qed