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

  r1, r2, r3, r4, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        int#_A(x1,x2) = max{4, x2 - 1}
        |0|_A = 5
        s_A(x1) = x1 + 16
        intlist_A(x1) = x1 + 34
        nil_A = 4
        int_A(x1,x2) = max{75, x1 + 3, x2 - 17}
        cons_A(x1,x2) = max{74, x1 - 9, x2 + 35}
  
    base order 2:
  
      precedence:
      
        intlist > cons > s = int > int# > |0| > nil
      
  
    partial status:
  
      pi(int#) = []
      pi(|0|) = []
      pi(s) = []
      pi(intlist) = [1]
      pi(nil) = []
      pi(int) = [1]
      pi(cons) = [2]

The following rules are removed:

  p2



-- 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

  r1, r2, r3, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        intlist#_A(x1) = max{4, x1 + 2}
        cons_A(x1,x2) = max{5, x1 - 11, x2 + 3}
        intlist_A(x1) = max{24, x1 - 16}
        nil_A = 7
        int_A(x1,x2) = max{23, x1 - 29, x2 - 10}
        s_A(x1) = max{6, x1 - 28}
        |0|_A = 8
  
    base order 2:
  
      precedence:
      
        intlist# = intlist > cons > s > int > nil > |0|
      
  
    partial status:
  
      pi(intlist#) = [1]
      pi(cons) = [2]
      pi(intlist) = []
      pi(nil) = []
      pi(int) = []
      pi(s) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.