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(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))
  inc() -> app(map(),app(plus(),app(s(),|0|())))

-- 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#(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)
p8: inc#() -> app#(map(),app(plus(),app(s(),|0|())))
p9: inc#() -> app#(plus(),app(s(),|0|()))
p10: inc#() -> app#(s(),|0|())

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(app(map(),f),nil()) -> nil()
r4: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r5: inc() -> app(map(),app(plus(),app(s(),|0|())))

The estimated dependency graph contains the following SCCs:

  {p6, p7}
  {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)

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(app(map(),f),nil()) -> nil()
r4: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r5: inc() -> app(map(),app(plus(),app(s(),|0|())))

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

The next rules are strictly ordered:

  p1, p2

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(app(map(),f),nil()) -> nil()
r4: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r5: inc() -> app(map(),app(plus(),app(s(),|0|())))

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

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5

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