YES

Problem:
 f(0(),n) -> g(0(),n)
 f(m,0()) -> g(m,0())
 f(s(m),s(n)) -> h(m,n,f(m,p(m,n)),f(s(m),n))
 g(n,m) -> bot()
 p(m,n) -> bot()
 h(k,l,m,n) -> bot()

Proof:
 DP Processor:
  DPs:
   f#(0(),n) -> g#(0(),n)
   f#(m,0()) -> g#(m,0())
   f#(s(m),s(n)) -> f#(s(m),n)
   f#(s(m),s(n)) -> p#(m,n)
   f#(s(m),s(n)) -> f#(m,p(m,n))
   f#(s(m),s(n)) -> h#(m,n,f(m,p(m,n)),f(s(m),n))
  TRS:
   f(0(),n) -> g(0(),n)
   f(m,0()) -> g(m,0())
   f(s(m),s(n)) -> h(m,n,f(m,p(m,n)),f(s(m),n))
   g(n,m) -> bot()
   p(m,n) -> bot()
   h(k,l,m,n) -> bot()
  KBO Processor:
   argument filtering:
    pi(0) = []
    pi(f) = 1
    pi(g) = 1
    pi(s) = [0]
    pi(p) = []
    pi(h) = 3
    pi(bot) = []
    pi(f#) = [0,1]
    pi(g#) = 1
    pi(p#) = []
    pi(h#) = [0,1]
   usable rules:
    p(m,n) -> bot()
   weight function:
    w0 = 1
    w(h#) = w(p#) = w(g#) = w(f#) = w(bot) = w(h) = w(p) = w(g) = w(0) = 1
    w(s) = w(f) = 0
   precedence:
    f# ~ p ~ s ~ 0 > h# ~ p# ~ g# ~ bot ~ h ~ g ~ f
   problem:
    DPs:
     
    TRS:
     f(0(),n) -> g(0(),n)
     f(m,0()) -> g(m,0())
     f(s(m),s(n)) -> h(m,n,f(m,p(m,n)),f(s(m),n))
     g(n,m) -> bot()
     p(m,n) -> bot()
     h(k,l,m,n) -> bot()
   Qed