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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      app#_A(x1,x2) = ((1,1),(0,0)) x1
      app_A(x1,x2) = ((1,1),(1,1)) x2 + (1,1)
      takeWhile_A() = (1,1)
      cons_A() = (1,1)
      dropWhile_A() = (1,1)

The next rules are strictly ordered:

  p1, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: app#(app(dropWhile(),p),app(app(cons(),x),xs)) -> app#(app(dropWhile(),p),xs)
p2: 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 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(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 set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      app#_A(x1,x2) = ((0,1),(1,1)) x2
      app_A(x1,x2) = ((0,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (0,1)
      dropWhile_A() = (0,0)
      cons_A() = (0,0)

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(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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      app#_A(x1,x2) = ((0,1),(1,1)) x2
      app_A(x1,x2) = ((0,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (0,1)
      takeWhile_A() = (0,0)
      cons_A() = (0,0)

The next rules are strictly ordered:

  p1

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