MAYBE

Problem:
 minus(x,0()) -> x
 minus(s(x),s(y)) -> minus(x,y)
 quot(0(),s(y)) -> 0()
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
 plus(0(),y) -> y
 plus(s(x),y) -> s(plus(x,y))
 minus(minus(x,y),z) -> minus(x,plus(y,z))

Proof:
 DP Processor:
  DPs:
   minus#(s(x),s(y)) -> minus#(x,y)
   quot#(s(x),s(y)) -> minus#(x,y)
   quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
   plus#(s(x),y) -> plus#(x,y)
   minus#(minus(x,y),z) -> plus#(y,z)
   minus#(minus(x,y),z) -> minus#(x,plus(y,z))
  TRS:
   minus(x,0()) -> x
   minus(s(x),s(y)) -> minus(x,y)
   quot(0(),s(y)) -> 0()
   quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
   plus(0(),y) -> y
   plus(s(x),y) -> s(plus(x,y))
   minus(minus(x,y),z) -> minus(x,plus(y,z))
  Usable Rule Processor:
   DPs:
    minus#(s(x),s(y)) -> minus#(x,y)
    quot#(s(x),s(y)) -> minus#(x,y)
    quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
    plus#(s(x),y) -> plus#(x,y)
    minus#(minus(x,y),z) -> plus#(y,z)
    minus#(minus(x,y),z) -> minus#(x,plus(y,z))
   TRS:
    minus(x,0()) -> x
    minus(s(x),s(y)) -> minus(x,y)
    minus(minus(x,y),z) -> minus(x,plus(y,z))
    plus(0(),y) -> y
    plus(s(x),y) -> s(plus(x,y))
   Open