YES

We show the termination of the TRS R:

  app(app(plus(),|0|()),y) -> y
  app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
  app(inc(),xs) -> app(app(map(),app(plus(),app(s(),|0|()))),xs)
  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(plus(),app(s(),x)),y) -> app#(s(),app(app(plus(),x),y))
p2: app#(app(plus(),app(s(),x)),y) -> app#(app(plus(),x),y)
p3: app#(app(plus(),app(s(),x)),y) -> app#(plus(),x)
p4: app#(inc(),xs) -> app#(app(map(),app(plus(),app(s(),|0|()))),xs)
p5: app#(inc(),xs) -> app#(map(),app(plus(),app(s(),|0|())))
p6: app#(inc(),xs) -> app#(plus(),app(s(),|0|()))
p7: app#(inc(),xs) -> app#(s(),|0|())
p8: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(cons(),app(f,x)),app(app(map(),f),xs))
p9: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(cons(),app(f,x))
p10: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(f,x)
p11: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)

and R consists of:

r1: app(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(inc(),xs) -> app(app(map(),app(plus(),app(s(),|0|()))),xs)
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:

  {p4, p10, p11}
  {p2}


-- 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)
p2: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)
p3: app#(inc(),xs) -> app#(app(map(),app(plus(),app(s(),|0|()))),xs)

and R consists of:

r1: app(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(inc(),xs) -> app(app(map(),app(plus(),app(s(),|0|()))),xs)
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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = x2 + (10,6)
          app_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,3)
          map_A() = (8,2)
          cons_A() = (6,1)
          inc_A() = (21,4)
          plus_A() = (1,5)
          s_A() = (1,5)
          |0|_A() = (1,1)
  
    precedence:
    
      app# = app = map = cons = inc = plus = s = |0|
    
    partial status:
  
      pi(app#) = [2]
      pi(app) = []
      pi(map) = []
      pi(cons) = []
      pi(inc) = []
      pi(plus) = []
      pi(s) = []
      pi(|0|) = []

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#(app(map(),f),xs)
p2: app#(inc(),xs) -> app#(app(map(),app(plus(),app(s(),|0|()))),xs)

and R consists of:

r1: app(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(inc(),xs) -> app(app(map(),app(plus(),app(s(),|0|()))),xs)
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#(app(map(),f),xs)

and R consists of:

r1: app(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(inc(),xs) -> app(app(map(),app(plus(),app(s(),|0|()))),xs)
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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,0)) x2 + (4,4)
          app_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (1,3)
          map_A() = (1,1)
          cons_A() = (3,2)
  
    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.

-- Reduction pair.

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

p1: app#(app(plus(),app(s(),x)),y) -> app#(app(plus(),x),y)

and R consists of:

r1: app(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(inc(),xs) -> app(app(map(),app(plus(),app(s(),|0|()))),xs)
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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (4,1)
          app_A(x1,x2) = x2 + (3,2)
          plus_A() = (1,1)
          s_A() = (2,3)
  
    precedence:
    
      app# = app = plus = s
    
    partial status:
  
      pi(app#) = [1]
      pi(app) = [2]
      pi(plus) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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