YES

Problem:
 le(0(),Y) -> true()
 le(s(X),0()) -> false()
 le(s(X),s(Y)) -> le(X,Y)
 minus(0(),Y) -> 0()
 minus(s(X),Y) -> ifMinus(le(s(X),Y),s(X),Y)
 ifMinus(true(),s(X),Y) -> 0()
 ifMinus(false(),s(X),Y) -> s(minus(X,Y))
 quot(0(),s(Y)) -> 0()
 quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))

Proof:
 DP Processor:
  DPs:
   le#(s(X),s(Y)) -> le#(X,Y)
   minus#(s(X),Y) -> le#(s(X),Y)
   minus#(s(X),Y) -> ifMinus#(le(s(X),Y),s(X),Y)
   ifMinus#(false(),s(X),Y) -> minus#(X,Y)
   quot#(s(X),s(Y)) -> minus#(X,Y)
   quot#(s(X),s(Y)) -> quot#(minus(X,Y),s(Y))
  TRS:
   le(0(),Y) -> true()
   le(s(X),0()) -> false()
   le(s(X),s(Y)) -> le(X,Y)
   minus(0(),Y) -> 0()
   minus(s(X),Y) -> ifMinus(le(s(X),Y),s(X),Y)
   ifMinus(true(),s(X),Y) -> 0()
   ifMinus(false(),s(X),Y) -> s(minus(X,Y))
   quot(0(),s(Y)) -> 0()
   quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
  Usable Rule Processor:
   DPs:
    le#(s(X),s(Y)) -> le#(X,Y)
    minus#(s(X),Y) -> le#(s(X),Y)
    minus#(s(X),Y) -> ifMinus#(le(s(X),Y),s(X),Y)
    ifMinus#(false(),s(X),Y) -> minus#(X,Y)
    quot#(s(X),s(Y)) -> minus#(X,Y)
    quot#(s(X),s(Y)) -> quot#(minus(X,Y),s(Y))
   TRS:
    le(s(X),0()) -> false()
    le(s(X),s(Y)) -> le(X,Y)
    le(0(),Y) -> true()
    minus(0(),Y) -> 0()
    minus(s(X),Y) -> ifMinus(le(s(X),Y),s(X),Y)
    ifMinus(true(),s(X),Y) -> 0()
    ifMinus(false(),s(X),Y) -> s(minus(X,Y))
   Matrix Interpretation Processor: dim=1
    
    usable rules:
     le(s(X),0()) -> false()
     le(s(X),s(Y)) -> le(X,Y)
     le(0(),Y) -> true()
     minus(0(),Y) -> 0()
     minus(s(X),Y) -> ifMinus(le(s(X),Y),s(X),Y)
     ifMinus(true(),s(X),Y) -> 0()
     ifMinus(false(),s(X),Y) -> s(minus(X,Y))
    interpretation:
     [quot#](x0, x1) = 3x0 + 6x1,
     
     [ifMinus#](x0, x1, x2) = x1,
     
     [minus#](x0, x1) = x0 + 2,
     
     [le#](x0, x1) = x0,
     
     [ifMinus](x0, x1, x2) = 2x1,
     
     [minus](x0, x1) = 2x0,
     
     [false] = 0,
     
     [s](x0) = 2x0 + 3,
     
     [true] = 0,
     
     [le](x0, x1) = x1,
     
     [0] = 0
    orientation:
     le#(s(X),s(Y)) = 2X + 3 >= X = le#(X,Y)
     
     minus#(s(X),Y) = 2X + 5 >= 2X + 3 = le#(s(X),Y)
     
     minus#(s(X),Y) = 2X + 5 >= 2X + 3 = ifMinus#(le(s(X),Y),s(X),Y)
     
     ifMinus#(false(),s(X),Y) = 2X + 3 >= X + 2 = minus#(X,Y)
     
     quot#(s(X),s(Y)) = 6X + 12Y + 27 >= X + 2 = minus#(X,Y)
     
     quot#(s(X),s(Y)) = 6X + 12Y + 27 >= 6X + 12Y + 18 = quot#(minus(X,Y),s(Y))
     
     le(s(X),0()) = 0 >= 0 = false()
     
     le(s(X),s(Y)) = 2Y + 3 >= Y = le(X,Y)
     
     le(0(),Y) = Y >= 0 = true()
     
     minus(0(),Y) = 0 >= 0 = 0()
     
     minus(s(X),Y) = 4X + 6 >= 4X + 6 = ifMinus(le(s(X),Y),s(X),Y)
     
     ifMinus(true(),s(X),Y) = 4X + 6 >= 0 = 0()
     
     ifMinus(false(),s(X),Y) = 4X + 6 >= 4X + 3 = s(minus(X,Y))
    problem:
     DPs:
      
     TRS:
      le(s(X),0()) -> false()
      le(s(X),s(Y)) -> le(X,Y)
      le(0(),Y) -> true()
      minus(0(),Y) -> 0()
      minus(s(X),Y) -> ifMinus(le(s(X),Y),s(X),Y)
      ifMinus(true(),s(X),Y) -> 0()
      ifMinus(false(),s(X),Y) -> s(minus(X,Y))
    Qed