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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          int#_A(x1,x2) = x1 + 1
          |0|_A() = 3
          s_A(x1) = x1
          intlist_A(x1) = 3
          nil_A() = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          int#_A(x1,x2) = max{0, x1 - 41, x2 - 42}
          |0|_A = 13
          s_A(x1) = max{37, x1 + 27}
          intlist_A(x1) = max{22, x1 - 25}
          nil_A = 1
          int#!_A(x1,x2) = max{1, x1 - 25, x2 - 21}
          |0|!_A = 7
          s!_A(x1) = max{17, x1 - 88}
          intlist!_A(x1) = max{18, x1 - 24}
          nil!_A = 6
  
    partial status:
  
      pi(int#) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(intlist) = []
      pi(nil) = []

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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          intlist#_A(x1) = x1
          cons_A(x1,x2) = x2 + 1
          intlist_A(x1) = 3
          nil_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          intlist#_A(x1) = 0
          cons_A(x1,x2) = max{2, x1 - 22, x2 - 6}
          intlist_A(x1) = max{5, x1 - 23}
          nil_A = 3
          intlist#!_A(x1) = 0
          cons!_A(x1,x2) = max{6, x1 - 21, x2 - 6}
          intlist!_A(x1) = max{5, x1 - 4}
          nil!_A = 4
  
    partial status:
  
      pi(intlist#) = []
      pi(cons) = [2]
      pi(intlist) = []
      pi(nil) = []

The following rules are removed:

  p1

Then no dependency pair remains.