YES

We show the termination of the TRS R:

  app(app(app(if(),true()),x),y) -> x
  app(app(app(if(),true()),x),y) -> y
  app(app(takeWhile(),p),nil()) -> nil()
  app(app(takeWhile(),p),app(app(cons(),x),xs)) -> app(app(app(if(),app(p,x)),app(app(cons(),x),app(app(takeWhile(),p),xs))),nil())
  app(app(dropWhile(),p),nil()) -> nil()
  app(app(dropWhile(),p),app(app(cons(),x),xs)) -> app(app(app(if(),app(p,x)),app(app(dropWhile(),p),xs)),app(app(cons(),x),xs))

-- SCC decomposition.

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

p1: app#(app(takeWhile(),p),app(app(cons(),x),xs)) -> app#(app(app(if(),app(p,x)),app(app(cons(),x),app(app(takeWhile(),p),xs))),nil())
p2: app#(app(takeWhile(),p),app(app(cons(),x),xs)) -> app#(app(if(),app(p,x)),app(app(cons(),x),app(app(takeWhile(),p),xs)))
p3: app#(app(takeWhile(),p),app(app(cons(),x),xs)) -> app#(if(),app(p,x))
p4: app#(app(takeWhile(),p),app(app(cons(),x),xs)) -> app#(p,x)
p5: app#(app(takeWhile(),p),app(app(cons(),x),xs)) -> app#(app(cons(),x),app(app(takeWhile(),p),xs))
p6: app#(app(takeWhile(),p),app(app(cons(),x),xs)) -> app#(app(takeWhile(),p),xs)
p7: app#(app(dropWhile(),p),app(app(cons(),x),xs)) -> app#(app(app(if(),app(p,x)),app(app(dropWhile(),p),xs)),app(app(cons(),x),xs))
p8: app#(app(dropWhile(),p),app(app(cons(),x),xs)) -> app#(app(if(),app(p,x)),app(app(dropWhile(),p),xs))
p9: app#(app(dropWhile(),p),app(app(cons(),x),xs)) -> app#(if(),app(p,x))
p10: app#(app(dropWhile(),p),app(app(cons(),x),xs)) -> app#(p,x)
p11: app#(app(dropWhile(),p),app(app(cons(),x),xs)) -> app#(app(dropWhile(),p),xs)

and R consists of:

r1: app(app(app(if(),true()),x),y) -> x
r2: app(app(app(if(),true()),x),y) -> y
r3: app(app(takeWhile(),p),nil()) -> nil()
r4: app(app(takeWhile(),p),app(app(cons(),x),xs)) -> app(app(app(if(),app(p,x)),app(app(cons(),x),app(app(takeWhile(),p),xs))),nil())
r5: app(app(dropWhile(),p),nil()) -> nil()
r6: app(app(dropWhile(),p),app(app(cons(),x),xs)) -> app(app(app(if(),app(p,x)),app(app(dropWhile(),p),xs)),app(app(cons(),x),xs))

The estimated dependency graph contains the following SCCs:

  {p4, p6, p10, p11}


-- Reduction pair.

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

p1: app#(app(takeWhile(),p),app(app(cons(),x),xs)) -> app#(p,x)
p2: app#(app(dropWhile(),p),app(app(cons(),x),xs)) -> app#(app(dropWhile(),p),xs)
p3: app#(app(dropWhile(),p),app(app(cons(),x),xs)) -> app#(p,x)
p4: app#(app(takeWhile(),p),app(app(cons(),x),xs)) -> app#(app(takeWhile(),p),xs)

and R consists of:

r1: app(app(app(if(),true()),x),y) -> x
r2: app(app(app(if(),true()),x),y) -> y
r3: app(app(takeWhile(),p),nil()) -> nil()
r4: app(app(takeWhile(),p),app(app(cons(),x),xs)) -> app(app(app(if(),app(p,x)),app(app(cons(),x),app(app(takeWhile(),p),xs))),nil())
r5: app(app(dropWhile(),p),nil()) -> nil()
r6: app(app(dropWhile(),p),app(app(cons(),x),xs)) -> app(app(app(if(),app(p,x)),app(app(dropWhile(),p),xs)),app(app(cons(),x),xs))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1 + x2
        app_A(x1,x2) = x1 + x2
        takeWhile_A() = 1
        cons_A() = 1
        dropWhile_A() = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        app > dropWhile > takeWhile > app# > cons
      
      argument filter:
    
        pi(app#) = []
        pi(app) = [1, 2]
        pi(takeWhile) = []
        pi(cons) = []
        pi(dropWhile) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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