YES

We show the termination of the TRS R:

  int(|0|(),|0|()) -> .(|0|(),nil())
  int(|0|(),s(y)) -> .(|0|(),int(s(|0|()),s(y)))
  int(s(x),|0|()) -> nil()
  int(s(x),s(y)) -> int_list(int(x,y))
  int_list(nil()) -> nil()
  int_list(.(x,y)) -> .(s(x),int_list(y))

-- SCC decomposition.

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_list#(int(x,y))
p3: int#(s(x),s(y)) -> int#(x,y)
p4: int_list#(.(x,y)) -> int_list#(y)

and R consists of:

r1: int(|0|(),|0|()) -> .(|0|(),nil())
r2: int(|0|(),s(y)) -> .(|0|(),int(s(|0|()),s(y)))
r3: int(s(x),|0|()) -> nil()
r4: int(s(x),s(y)) -> int_list(int(x,y))
r5: int_list(nil()) -> nil()
r6: int_list(.(x,y)) -> .(s(x),int_list(y))

The estimated dependency graph contains the following SCCs:

  {p1, p3}
  {p4}


-- 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: int(|0|(),|0|()) -> .(|0|(),nil())
r2: int(|0|(),s(y)) -> .(|0|(),int(s(|0|()),s(y)))
r3: int(s(x),|0|()) -> nil()
r4: int(s(x),s(y)) -> int_list(int(x,y))
r5: int_list(nil()) -> nil()
r6: int_list(.(x,y)) -> .(s(x),int_list(y))

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: int(|0|(),|0|()) -> .(|0|(),nil())
r2: int(|0|(),s(y)) -> .(|0|(),int(s(|0|()),s(y)))
r3: int(s(x),|0|()) -> nil()
r4: int(s(x),s(y)) -> int_list(int(x,y))
r5: int_list(nil()) -> nil()
r6: int_list(.(x,y)) -> .(s(x),int_list(y))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: int_list#(.(x,y)) -> int_list#(y)

and R consists of:

r1: int(|0|(),|0|()) -> .(|0|(),nil())
r2: int(|0|(),s(y)) -> .(|0|(),int(s(|0|()),s(y)))
r3: int(s(x),|0|()) -> nil()
r4: int(s(x),s(y)) -> int_list(int(x,y))
r5: int_list(nil()) -> nil()
r6: int_list(.(x,y)) -> .(s(x),int_list(y))

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:
      
        . > int_list#
      
      argument filter:
    
        pi(int_list#) = [1]
        pi(.) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        . > int_list#
      
      argument filter:
    
        pi(int_list#) = [1]
        pi(.) = [1, 2]
    

The next rules are strictly ordered:

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

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