YES

Problem:
 pred(s(x)) -> x
 minus(x,0()) -> x
 minus(x,s(y)) -> pred(minus(x,y))
 quot(0(),s(y)) -> 0()
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
 log(s(0())) -> 0()
 log(s(s(x))) -> s(log(s(quot(x,s(s(0()))))))

Proof:
 DP Processor:
  DPs:
   minus#(x,s(y)) -> minus#(x,y)
   minus#(x,s(y)) -> pred#(minus(x,y))
   quot#(s(x),s(y)) -> minus#(x,y)
   quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
   log#(s(s(x))) -> quot#(x,s(s(0())))
   log#(s(s(x))) -> log#(s(quot(x,s(s(0())))))
  TRS:
   pred(s(x)) -> x
   minus(x,0()) -> x
   minus(x,s(y)) -> pred(minus(x,y))
   quot(0(),s(y)) -> 0()
   quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
   log(s(0())) -> 0()
   log(s(s(x))) -> s(log(s(quot(x,s(s(0()))))))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [log#](x0) = x0 + 1,
    
    [quot#](x0, x1) = x0 + x1,
    
    [minus#](x0, x1) = x0 + x1,
    
    [pred#](x0) = 0,
    
    [log](x0) = 4x0,
    
    [quot](x0, x1) = x0,
    
    [minus](x0, x1) = x0,
    
    [0] = 0,
    
    [pred](x0) = x0,
    
    [s](x0) = x0 + 1
   orientation:
    minus#(x,s(y)) = x + y + 1 >= x + y = minus#(x,y)
    
    minus#(x,s(y)) = x + y + 1 >= 0 = pred#(minus(x,y))
    
    quot#(s(x),s(y)) = x + y + 2 >= x + y = minus#(x,y)
    
    quot#(s(x),s(y)) = x + y + 2 >= x + y + 1 = quot#(minus(x,y),s(y))
    
    log#(s(s(x))) = x + 3 >= x + 2 = quot#(x,s(s(0())))
    
    log#(s(s(x))) = x + 3 >= x + 2 = log#(s(quot(x,s(s(0())))))
    
    pred(s(x)) = x + 1 >= x = x
    
    minus(x,0()) = x >= x = x
    
    minus(x,s(y)) = x >= x = pred(minus(x,y))
    
    quot(0(),s(y)) = 0 >= 0 = 0()
    
    quot(s(x),s(y)) = x + 1 >= x + 1 = s(quot(minus(x,y),s(y)))
    
    log(s(0())) = 4 >= 0 = 0()
    
    log(s(s(x))) = 4x + 8 >= 4x + 5 = s(log(s(quot(x,s(s(0()))))))
   problem:
    DPs:
     
    TRS:
     pred(s(x)) -> x
     minus(x,0()) -> x
     minus(x,s(y)) -> pred(minus(x,y))
     quot(0(),s(y)) -> 0()
     quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
     log(s(0())) -> 0()
     log(s(s(x))) -> s(log(s(quot(x,s(s(0()))))))
   Qed