MAYBE

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs:
  { a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N)))
  , a__terms(X) -> terms(X)
  , a__sqr(X) -> sqr(X)
  , a__sqr(s(X)) -> s(add(sqr(X), dbl(X)))
  , a__sqr(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(recip(X)) -> recip(mark(X))
  , mark(terms(X)) -> a__terms(mark(X))
  , mark(s(X)) -> s(X)
  , mark(0()) -> 0()
  , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
  , mark(sqr(X)) -> a__sqr(mark(X))
  , mark(dbl(X)) -> a__dbl(mark(X))
  , mark(nil()) -> nil()
  , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
  , a__dbl(X) -> dbl(X)
  , a__dbl(s(X)) -> s(s(dbl(X)))
  , a__dbl(0()) -> 0()
  , a__add(X1, X2) -> add(X1, X2)
  , a__add(s(X), Y) -> s(add(X, Y))
  , a__add(0(), X) -> mark(X)
  , a__first(X1, X2) -> first(X1, X2)
  , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
  , a__first(0(), X) -> nil() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { a__terms^#(N) -> c_1(a__sqr^#(mark(N)), mark^#(N))
  , a__terms^#(X) -> c_2()
  , a__sqr^#(X) -> c_3()
  , a__sqr^#(s(X)) -> c_4()
  , a__sqr^#(0()) -> c_5()
  , mark^#(cons(X1, X2)) -> c_6(mark^#(X1))
  , mark^#(recip(X)) -> c_7(mark^#(X))
  , mark^#(terms(X)) -> c_8(a__terms^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_9()
  , mark^#(0()) -> c_10()
  , mark^#(add(X1, X2)) ->
    c_11(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(sqr(X)) -> c_12(a__sqr^#(mark(X)), mark^#(X))
  , mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X)), mark^#(X))
  , mark^#(nil()) -> c_14()
  , mark^#(first(X1, X2)) ->
    c_15(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__add^#(X1, X2) -> c_19()
  , a__add^#(s(X), Y) -> c_20()
  , a__add^#(0(), X) -> c_21(mark^#(X))
  , a__dbl^#(X) -> c_16()
  , a__dbl^#(s(X)) -> c_17()
  , a__dbl^#(0()) -> c_18()
  , a__first^#(X1, X2) -> c_22()
  , a__first^#(s(X), cons(Y, Z)) -> c_23(mark^#(Y))
  , a__first^#(0(), X) -> c_24() }

and mark the set of starting terms.

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { a__terms^#(N) -> c_1(a__sqr^#(mark(N)), mark^#(N))
  , a__terms^#(X) -> c_2()
  , a__sqr^#(X) -> c_3()
  , a__sqr^#(s(X)) -> c_4()
  , a__sqr^#(0()) -> c_5()
  , mark^#(cons(X1, X2)) -> c_6(mark^#(X1))
  , mark^#(recip(X)) -> c_7(mark^#(X))
  , mark^#(terms(X)) -> c_8(a__terms^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_9()
  , mark^#(0()) -> c_10()
  , mark^#(add(X1, X2)) ->
    c_11(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(sqr(X)) -> c_12(a__sqr^#(mark(X)), mark^#(X))
  , mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X)), mark^#(X))
  , mark^#(nil()) -> c_14()
  , mark^#(first(X1, X2)) ->
    c_15(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__add^#(X1, X2) -> c_19()
  , a__add^#(s(X), Y) -> c_20()
  , a__add^#(0(), X) -> c_21(mark^#(X))
  , a__dbl^#(X) -> c_16()
  , a__dbl^#(s(X)) -> c_17()
  , a__dbl^#(0()) -> c_18()
  , a__first^#(X1, X2) -> c_22()
  , a__first^#(s(X), cons(Y, Z)) -> c_23(mark^#(Y))
  , a__first^#(0(), X) -> c_24() }
Weak Trs:
  { a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N)))
  , a__terms(X) -> terms(X)
  , a__sqr(X) -> sqr(X)
  , a__sqr(s(X)) -> s(add(sqr(X), dbl(X)))
  , a__sqr(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(recip(X)) -> recip(mark(X))
  , mark(terms(X)) -> a__terms(mark(X))
  , mark(s(X)) -> s(X)
  , mark(0()) -> 0()
  , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
  , mark(sqr(X)) -> a__sqr(mark(X))
  , mark(dbl(X)) -> a__dbl(mark(X))
  , mark(nil()) -> nil()
  , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
  , a__dbl(X) -> dbl(X)
  , a__dbl(s(X)) -> s(s(dbl(X)))
  , a__dbl(0()) -> 0()
  , a__add(X1, X2) -> add(X1, X2)
  , a__add(s(X), Y) -> s(add(X, Y))
  , a__add(0(), X) -> mark(X)
  , a__first(X1, X2) -> first(X1, X2)
  , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
  , a__first(0(), X) -> nil() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of
{2,3,4,5,9,10,14,16,17,19,20,21,22,24} by applications of
Pre({2,3,4,5,9,10,14,16,17,19,20,21,22,24}) =
{1,6,7,8,11,12,13,15,18,23}. Here rules are labeled as follows:

  DPs:
    { 1: a__terms^#(N) -> c_1(a__sqr^#(mark(N)), mark^#(N))
    , 2: a__terms^#(X) -> c_2()
    , 3: a__sqr^#(X) -> c_3()
    , 4: a__sqr^#(s(X)) -> c_4()
    , 5: a__sqr^#(0()) -> c_5()
    , 6: mark^#(cons(X1, X2)) -> c_6(mark^#(X1))
    , 7: mark^#(recip(X)) -> c_7(mark^#(X))
    , 8: mark^#(terms(X)) -> c_8(a__terms^#(mark(X)), mark^#(X))
    , 9: mark^#(s(X)) -> c_9()
    , 10: mark^#(0()) -> c_10()
    , 11: mark^#(add(X1, X2)) ->
          c_11(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 12: mark^#(sqr(X)) -> c_12(a__sqr^#(mark(X)), mark^#(X))
    , 13: mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X)), mark^#(X))
    , 14: mark^#(nil()) -> c_14()
    , 15: mark^#(first(X1, X2)) ->
          c_15(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 16: a__add^#(X1, X2) -> c_19()
    , 17: a__add^#(s(X), Y) -> c_20()
    , 18: a__add^#(0(), X) -> c_21(mark^#(X))
    , 19: a__dbl^#(X) -> c_16()
    , 20: a__dbl^#(s(X)) -> c_17()
    , 21: a__dbl^#(0()) -> c_18()
    , 22: a__first^#(X1, X2) -> c_22()
    , 23: a__first^#(s(X), cons(Y, Z)) -> c_23(mark^#(Y))
    , 24: a__first^#(0(), X) -> c_24() }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { a__terms^#(N) -> c_1(a__sqr^#(mark(N)), mark^#(N))
  , mark^#(cons(X1, X2)) -> c_6(mark^#(X1))
  , mark^#(recip(X)) -> c_7(mark^#(X))
  , mark^#(terms(X)) -> c_8(a__terms^#(mark(X)), mark^#(X))
  , mark^#(add(X1, X2)) ->
    c_11(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(sqr(X)) -> c_12(a__sqr^#(mark(X)), mark^#(X))
  , mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X)), mark^#(X))
  , mark^#(first(X1, X2)) ->
    c_15(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__add^#(0(), X) -> c_21(mark^#(X))
  , a__first^#(s(X), cons(Y, Z)) -> c_23(mark^#(Y)) }
Weak DPs:
  { a__terms^#(X) -> c_2()
  , a__sqr^#(X) -> c_3()
  , a__sqr^#(s(X)) -> c_4()
  , a__sqr^#(0()) -> c_5()
  , mark^#(s(X)) -> c_9()
  , mark^#(0()) -> c_10()
  , mark^#(nil()) -> c_14()
  , a__add^#(X1, X2) -> c_19()
  , a__add^#(s(X), Y) -> c_20()
  , a__dbl^#(X) -> c_16()
  , a__dbl^#(s(X)) -> c_17()
  , a__dbl^#(0()) -> c_18()
  , a__first^#(X1, X2) -> c_22()
  , a__first^#(0(), X) -> c_24() }
Weak Trs:
  { a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N)))
  , a__terms(X) -> terms(X)
  , a__sqr(X) -> sqr(X)
  , a__sqr(s(X)) -> s(add(sqr(X), dbl(X)))
  , a__sqr(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(recip(X)) -> recip(mark(X))
  , mark(terms(X)) -> a__terms(mark(X))
  , mark(s(X)) -> s(X)
  , mark(0()) -> 0()
  , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
  , mark(sqr(X)) -> a__sqr(mark(X))
  , mark(dbl(X)) -> a__dbl(mark(X))
  , mark(nil()) -> nil()
  , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
  , a__dbl(X) -> dbl(X)
  , a__dbl(s(X)) -> s(s(dbl(X)))
  , a__dbl(0()) -> 0()
  , a__add(X1, X2) -> add(X1, X2)
  , a__add(s(X), Y) -> s(add(X, Y))
  , a__add(0(), X) -> mark(X)
  , a__first(X1, X2) -> first(X1, X2)
  , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
  , a__first(0(), X) -> nil() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ a__terms^#(X) -> c_2()
, a__sqr^#(X) -> c_3()
, a__sqr^#(s(X)) -> c_4()
, a__sqr^#(0()) -> c_5()
, mark^#(s(X)) -> c_9()
, mark^#(0()) -> c_10()
, mark^#(nil()) -> c_14()
, a__add^#(X1, X2) -> c_19()
, a__add^#(s(X), Y) -> c_20()
, a__dbl^#(X) -> c_16()
, a__dbl^#(s(X)) -> c_17()
, a__dbl^#(0()) -> c_18()
, a__first^#(X1, X2) -> c_22()
, a__first^#(0(), X) -> c_24() }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { a__terms^#(N) -> c_1(a__sqr^#(mark(N)), mark^#(N))
  , mark^#(cons(X1, X2)) -> c_6(mark^#(X1))
  , mark^#(recip(X)) -> c_7(mark^#(X))
  , mark^#(terms(X)) -> c_8(a__terms^#(mark(X)), mark^#(X))
  , mark^#(add(X1, X2)) ->
    c_11(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(sqr(X)) -> c_12(a__sqr^#(mark(X)), mark^#(X))
  , mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X)), mark^#(X))
  , mark^#(first(X1, X2)) ->
    c_15(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__add^#(0(), X) -> c_21(mark^#(X))
  , a__first^#(s(X), cons(Y, Z)) -> c_23(mark^#(Y)) }
Weak Trs:
  { a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N)))
  , a__terms(X) -> terms(X)
  , a__sqr(X) -> sqr(X)
  , a__sqr(s(X)) -> s(add(sqr(X), dbl(X)))
  , a__sqr(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(recip(X)) -> recip(mark(X))
  , mark(terms(X)) -> a__terms(mark(X))
  , mark(s(X)) -> s(X)
  , mark(0()) -> 0()
  , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
  , mark(sqr(X)) -> a__sqr(mark(X))
  , mark(dbl(X)) -> a__dbl(mark(X))
  , mark(nil()) -> nil()
  , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
  , a__dbl(X) -> dbl(X)
  , a__dbl(s(X)) -> s(s(dbl(X)))
  , a__dbl(0()) -> 0()
  , a__add(X1, X2) -> add(X1, X2)
  , a__add(s(X), Y) -> s(add(X, Y))
  , a__add(0(), X) -> mark(X)
  , a__first(X1, X2) -> first(X1, X2)
  , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
  , a__first(0(), X) -> nil() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

  { a__terms^#(N) -> c_1(a__sqr^#(mark(N)), mark^#(N))
  , mark^#(sqr(X)) -> c_12(a__sqr^#(mark(X)), mark^#(X))
  , mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X)), mark^#(X)) }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { a__terms^#(N) -> c_1(mark^#(N))
  , mark^#(cons(X1, X2)) -> c_2(mark^#(X1))
  , mark^#(recip(X)) -> c_3(mark^#(X))
  , mark^#(terms(X)) -> c_4(a__terms^#(mark(X)), mark^#(X))
  , mark^#(add(X1, X2)) ->
    c_5(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(sqr(X)) -> c_6(mark^#(X))
  , mark^#(dbl(X)) -> c_7(mark^#(X))
  , mark^#(first(X1, X2)) ->
    c_8(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__add^#(0(), X) -> c_9(mark^#(X))
  , a__first^#(s(X), cons(Y, Z)) -> c_10(mark^#(Y)) }
Weak Trs:
  { a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N)))
  , a__terms(X) -> terms(X)
  , a__sqr(X) -> sqr(X)
  , a__sqr(s(X)) -> s(add(sqr(X), dbl(X)))
  , a__sqr(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(recip(X)) -> recip(mark(X))
  , mark(terms(X)) -> a__terms(mark(X))
  , mark(s(X)) -> s(X)
  , mark(0()) -> 0()
  , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
  , mark(sqr(X)) -> a__sqr(mark(X))
  , mark(dbl(X)) -> a__dbl(mark(X))
  , mark(nil()) -> nil()
  , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
  , a__dbl(X) -> dbl(X)
  , a__dbl(s(X)) -> s(s(dbl(X)))
  , a__dbl(0()) -> 0()
  , a__add(X1, X2) -> add(X1, X2)
  , a__add(s(X), Y) -> s(add(X, Y))
  , a__add(0(), X) -> mark(X)
  , a__first(X1, X2) -> first(X1, X2)
  , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
  , a__first(0(), X) -> nil() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..