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
  Usable Rule Processor:
   DPs:
    p#(m,n,s(r)) -> p#(m,r,n)
    p#(m,s(n),0()) -> p#(0(),n,m)
   TRS:
    
   KBO Processor:
    argument filtering:
     pi(s) = [0]
     pi(0) = []
     pi(p#) = [0,1,2]
    usable rules:
     
    weight function:
     w0 = 1
     w(0) = w(s) = 1
     w(p#) = 0
    precedence:
     p# ~ 0 ~ s
    problem:
     DPs:
      
     TRS:
      
    Qed