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
  CDG 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
   graph:
    
   Qed