YES

We show the termination of the TRS R:

  app(id(),x) -> x
  app(add(),|0|()) -> id()
  app(app(add(),app(s(),x)),y) -> app(s(),app(app(add(),x),y))
  app(app(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))

-- SCC decomposition.

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

p1: app#(app(add(),app(s(),x)),y) -> app#(s(),app(app(add(),x),y))
p2: app#(app(add(),app(s(),x)),y) -> app#(app(add(),x),y)
p3: app#(app(add(),app(s(),x)),y) -> app#(add(),x)
p4: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(cons(),app(f,x)),app(app(map(),f),xs))
p5: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(cons(),app(f,x))
p6: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(f,x)
p7: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)

and R consists of:

r1: app(id(),x) -> x
r2: app(add(),|0|()) -> id()
r3: app(app(add(),app(s(),x)),y) -> app(s(),app(app(add(),x),y))
r4: app(app(map(),f),nil()) -> nil()
r5: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))

The estimated dependency graph contains the following SCCs:

  {p2, p6, p7}


-- Reduction pair.

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

p1: app#(app(add(),app(s(),x)),y) -> app#(app(add(),x),y)
p2: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)
p3: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(f,x)

and R consists of:

r1: app(id(),x) -> x
r2: app(add(),|0|()) -> id()
r3: app(app(add(),app(s(),x)),y) -> app(s(),app(app(add(),x),y))
r4: app(app(map(),f),nil()) -> nil()
r5: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))

The set of usable rules consists of

  r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        app > id > |0| > map > app# > cons > add > s
      
      argument filter:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(add) = []
        pi(s) = []
        pi(map) = []
        pi(cons) = []
        pi(|0|) = []
        pi(id) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        id > |0| > app > map > app# > cons > add > s
      
      argument filter:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(add) = []
        pi(s) = []
        pi(map) = []
        pi(cons) = []
        pi(|0|) = []
        pi(id) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        id > |0| > app > map > app# > cons > add > s
      
      argument filter:
    
        pi(app#) = 2
        pi(app) = [1, 2]
        pi(add) = []
        pi(s) = []
        pi(map) = []
        pi(cons) = []
        pi(|0|) = []
        pi(id) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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