YES

Problem:
 perfectp(0()) -> false()
 perfectp(s(x)) -> f(x,s(0()),s(x),s(x))
 f(0(),y,0(),u) -> true()
 f(0(),y,s(z),u) -> false()
 f(s(x),0(),z,u) -> f(x,u,minus(z,s(x)),u)
 f(s(x),s(y),z,u) -> if(le(x,y),f(s(x),minus(y,x),z,u),f(x,u,z,u))

Proof:
 DP Processor:
  DPs:
   perfectp#(s(x)) -> f#(x,s(0()),s(x),s(x))
   f#(s(x),0(),z,u) -> f#(x,u,minus(z,s(x)),u)
   f#(s(x),s(y),z,u) -> f#(x,u,z,u)
   f#(s(x),s(y),z,u) -> f#(s(x),minus(y,x),z,u)
  TRS:
   perfectp(0()) -> false()
   perfectp(s(x)) -> f(x,s(0()),s(x),s(x))
   f(0(),y,0(),u) -> true()
   f(0(),y,s(z),u) -> false()
   f(s(x),0(),z,u) -> f(x,u,minus(z,s(x)),u)
   f(s(x),s(y),z,u) -> if(le(x,y),f(s(x),minus(y,x),z,u),f(x,u,z,u))
  Usable Rule Processor:
   DPs:
    perfectp#(s(x)) -> f#(x,s(0()),s(x),s(x))
    f#(s(x),0(),z,u) -> f#(x,u,minus(z,s(x)),u)
    f#(s(x),s(y),z,u) -> f#(x,u,z,u)
    f#(s(x),s(y),z,u) -> f#(s(x),minus(y,x),z,u)
   TRS:
    
   LPO Processor:
    argument filtering:
     pi(0) = []
     pi(s) = [0]
     pi(minus) = []
     pi(perfectp#) = [0]
     pi(f#) = [0,1,3]
    usable rules:
     
    precedence:
     perfectp# > s > f# ~ minus ~ 0
    problem:
     DPs:
      
     TRS:
      
    Qed