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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = x2 + (8,5)
            app_A(x1,x2) = x1 + x2 + (2,2)
            add_A() = (1,6)
            s_A() = (2,7)
            map_A() = (5,1)
            cons_A() = (9,4)
            |0|_A() = (1,1)
            id_A() = (0,0)
    
      precedence:
      
        app# > s > app = add > map > cons = |0| = id
      
      partial status:
    
        pi(app#) = []
        pi(app) = [2]
        pi(add) = []
        pi(s) = []
        pi(map) = []
        pi(cons) = []
        pi(|0|) = []
        pi(id) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (0,0)
            app_A(x1,x2) = (0,0)
            add_A() = (1,1)
            s_A() = (2,2)
            map_A() = (1,1)
            cons_A() = (2,2)
            |0|_A() = (3,3)
            id_A() = (2,2)
    
      precedence:
      
        id > s > map > cons > |0| > app# = app = add
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(add) = []
        pi(s) = []
        pi(map) = []
        pi(cons) = []
        pi(|0|) = []
        pi(id) = []
    

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: app#(app(add(),app(s(),x)),y) -> app#(app(add(),x),y)
p2: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- 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#(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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = ((1,0),(0,0)) x1 + (3,3)
            app_A(x1,x2) = x2 + (2,1)
            add_A() = (3,4)
            s_A() = (4,5)
            map_A() = (1,1)
            cons_A() = (1,4)
            |0|_A() = (1,1)
            id_A() = (0,0)
    
      precedence:
      
        app# = app > add = s = map = cons = id > |0|
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(add) = []
        pi(s) = []
        pi(map) = []
        pi(cons) = []
        pi(|0|) = []
        pi(id) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (3,4)
            app_A(x1,x2) = (0,3)
            add_A() = (1,1)
            s_A() = (2,2)
            map_A() = (1,1)
            cons_A() = (1,1)
            |0|_A() = (1,3)
            id_A() = (0,2)
    
      precedence:
      
        cons > add = s = id > app# = map > app = |0|
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(add) = []
        pi(s) = []
        pi(map) = []
        pi(cons) = []
        pi(|0|) = []
        pi(id) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = ((1,0),(1,1)) x1 + (4,4)
            app_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (2,2)
            map_A() = (5,1)
            cons_A() = (1,1)
    
      precedence:
      
        app# = app = map = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(map) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (1,2)
            app_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (1,3)
            map_A() = (2,1)
            cons_A() = (2,3)
    
      precedence:
      
        app# = app > map = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(map) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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