YES

We show the termination of the TRS R:

  intlist(nil()) -> nil()
  int(s(x),|0|()) -> nil()
  int(x,x) -> cons(x,nil())
  intlist(cons(x,y)) -> cons(s(x),intlist(y))
  int(s(x),s(y)) -> intlist(int(x,y))
  int(|0|(),s(y)) -> cons(|0|(),int(s(|0|()),s(y)))
  intlist(cons(x,nil())) -> cons(s(x),nil())

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: intlist#(cons(x,y)) -> intlist#(y)
p2: int#(s(x),s(y)) -> intlist#(int(x,y))
p3: int#(s(x),s(y)) -> int#(x,y)
p4: int#(|0|(),s(y)) -> int#(s(|0|()),s(y))

and R consists of:

r1: intlist(nil()) -> nil()
r2: int(s(x),|0|()) -> nil()
r3: int(x,x) -> cons(x,nil())
r4: intlist(cons(x,y)) -> cons(s(x),intlist(y))
r5: int(s(x),s(y)) -> intlist(int(x,y))
r6: int(|0|(),s(y)) -> cons(|0|(),int(s(|0|()),s(y)))
r7: intlist(cons(x,nil())) -> cons(s(x),nil())

The estimated dependency graph contains the following SCCs:

  {p3, p4}
  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: int#(|0|(),s(y)) -> int#(s(|0|()),s(y))
p2: int#(s(x),s(y)) -> int#(x,y)

and R consists of:

r1: intlist(nil()) -> nil()
r2: int(s(x),|0|()) -> nil()
r3: int(x,x) -> cons(x,nil())
r4: intlist(cons(x,y)) -> cons(s(x),intlist(y))
r5: int(s(x),s(y)) -> intlist(int(x,y))
r6: int(|0|(),s(y)) -> cons(|0|(),int(s(|0|()),s(y)))
r7: intlist(cons(x,nil())) -> cons(s(x),nil())

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > int# > s
      
      argument filter:
    
        pi(int#) = [1, 2]
        pi(|0|) = []
        pi(s) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |0| > s > int#
      
      argument filter:
    
        pi(int#) = [2]
        pi(|0|) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: int#(|0|(),s(y)) -> int#(s(|0|()),s(y))

and R consists of:

r1: intlist(nil()) -> nil()
r2: int(s(x),|0|()) -> nil()
r3: int(x,x) -> cons(x,nil())
r4: intlist(cons(x,y)) -> cons(s(x),intlist(y))
r5: int(s(x),s(y)) -> intlist(int(x,y))
r6: int(|0|(),s(y)) -> cons(|0|(),int(s(|0|()),s(y)))
r7: intlist(cons(x,nil())) -> cons(s(x),nil())

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: intlist#(cons(x,y)) -> intlist#(y)

and R consists of:

r1: intlist(nil()) -> nil()
r2: int(s(x),|0|()) -> nil()
r3: int(x,x) -> cons(x,nil())
r4: intlist(cons(x,y)) -> cons(s(x),intlist(y))
r5: int(s(x),s(y)) -> intlist(int(x,y))
r6: int(|0|(),s(y)) -> cons(|0|(),int(s(|0|()),s(y)))
r7: intlist(cons(x,nil())) -> cons(s(x),nil())

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        cons > intlist#
      
      argument filter:
    
        pi(intlist#) = [1]
        pi(cons) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        cons > intlist#
      
      argument filter:
    
        pi(intlist#) = [1]
        pi(cons) = [1, 2]
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7

We remove them from the problem.  Then no dependency pair remains.