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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            int#_A(x1,x2) = max{13, x1 - 17}
            |0|_A = 2
            s_A(x1) = max{30, x1 + 16}
            int#!_A(x1,x2) = max{10, x1 - 19, x2 - 16}
            |0|!_A = 5
            s!_A(x1) = x1 + 9
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              int#_A(x1,x2) = x2
              |0|_A() = 1
              s_A(x1) = x1 + 2
              int#!_A(x1,x2) = x2 + 3
              |0|!_A() = 2
              s!_A(x1) = 4
        
  
    partial status:
  
      pi(int#) = [2]
      pi(|0|) = []
      pi(s) = [1]

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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            intlist#_A(x1) = 0
            cons_A(x1,x2) = max{20, x1 - 16, x2 - 6}
            intlist#!_A(x1) = max{1, x1 - 21}
            cons!_A(x1,x2) = max{6, x1 - 2, x2 - 4}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          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) = x1 + 5
              cons!_A(x1,x2) = 4
        
  
    partial status:
  
      pi(intlist#) = [1]
      pi(cons) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.