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)))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [quot#](x0, x1) = x0,
    
    [ifMinus#](x0, x1, x2) = 1/2x1,
    
    [minus#](x0, x1) = 1/2x0 + 1/2,
    
    [le#](x0, x1) = 1/2x0,
    
    [quot](x0, x1) = 3x0,
    
    [ifMinus](x0, x1, x2) = x1,
    
    [minus](x0, x1) = x0,
    
    [false] = 0,
    
    [s](x0) = x0 + 2,
    
    [true] = 0,
    
    [le](x0, x1) = 0,
    
    [0] = 0
   orientation:
    le#(s(X),s(Y)) = 1/2X + 1 >= 1/2X = le#(X,Y)
    
    minus#(s(X),Y) = 1/2X + 3/2 >= 1/2X + 1 = le#(s(X),Y)
    
    minus#(s(X),Y) = 1/2X + 3/2 >= 1/2X + 1 = ifMinus#(le(s(X),Y),s(X),Y)
    
    ifMinus#(false(),s(X),Y) = 1/2X + 1 >= 1/2X + 1/2 = minus#(X,Y)
    
    quot#(s(X),s(Y)) = X + 2 >= 1/2X + 1/2 = minus#(X,Y)
    
    quot#(s(X),s(Y)) = X + 2 >= X = quot#(minus(X,Y),s(Y))
    
    le(0(),Y) = 0 >= 0 = true()
    
    le(s(X),0()) = 0 >= 0 = false()
    
    le(s(X),s(Y)) = 0 >= 0 = le(X,Y)
    
    minus(0(),Y) = 0 >= 0 = 0()
    
    minus(s(X),Y) = X + 2 >= X + 2 = ifMinus(le(s(X),Y),s(X),Y)
    
    ifMinus(true(),s(X),Y) = X + 2 >= 0 = 0()
    
    ifMinus(false(),s(X),Y) = X + 2 >= X + 2 = s(minus(X,Y))
    
    quot(0(),s(Y)) = 0 >= 0 = 0()
    
    quot(s(X),s(Y)) = 3X + 6 >= 3X + 2 = s(quot(minus(X,Y),s(Y)))
   problem:
    DPs:
     
    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)))
   Qed