MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { pairNs() -> cons(0(), n__incr(oddNs())) , cons(X1, X2) -> n__cons(X1, X2) , oddNs() -> incr(pairNs()) , incr(X) -> n__incr(X) , incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) , activate(X) -> X , activate(n__incr(X)) -> incr(X) , activate(n__take(X1, X2)) -> take(X1, X2) , activate(n__zip(X1, X2)) -> zip(X1, X2) , activate(n__cons(X1, X2)) -> cons(X1, X2) , activate(n__repItems(X)) -> repItems(X) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) , zip(X1, X2) -> n__zip(X1, X2) , zip(X, nil()) -> nil() , zip(cons(X, XS), cons(Y, YS)) -> cons(pair(X, Y), n__zip(activate(XS), activate(YS))) , zip(nil(), XS) -> nil() , tail(cons(X, XS)) -> activate(XS) , repItems(X) -> n__repItems(X) , repItems(cons(X, XS)) -> cons(X, n__cons(X, n__repItems(activate(XS)))) , repItems(nil()) -> nil() } Obligation: innermost runtime complexity Answer: MAYBE Arguments of following rules are not normal-forms: { incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) , zip(cons(X, XS), cons(Y, YS)) -> cons(pair(X, Y), n__zip(activate(XS), activate(YS))) , tail(cons(X, XS)) -> activate(XS) , repItems(cons(X, XS)) -> cons(X, n__cons(X, n__repItems(activate(XS)))) } All above mentioned rules can be savely removed. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { pairNs() -> cons(0(), n__incr(oddNs())) , cons(X1, X2) -> n__cons(X1, X2) , oddNs() -> incr(pairNs()) , incr(X) -> n__incr(X) , activate(X) -> X , activate(n__incr(X)) -> incr(X) , activate(n__take(X1, X2)) -> take(X1, X2) , activate(n__zip(X1, X2)) -> zip(X1, X2) , activate(n__cons(X1, X2)) -> cons(X1, X2) , activate(n__repItems(X)) -> repItems(X) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , zip(X1, X2) -> n__zip(X1, X2) , zip(X, nil()) -> nil() , zip(nil(), XS) -> nil() , repItems(X) -> n__repItems(X) , repItems(nil()) -> nil() } Obligation: innermost runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'Best' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'Best' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'WithProblem (timeout of 30 seconds) (timeout of 60 seconds)' failed due to the following reason: Computation stopped due to timeout after 30.0 seconds. 2) 'Fastest (timeout of 5 seconds) (timeout of 60 seconds)' failed due to the following reason: Computation stopped due to timeout after 5.0 seconds. 3) 'Best' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'bsearch-popstar (timeout of 60 seconds)' failed due to the following reason: The input cannot be shown compatible 2) 'Polynomial Path Order (PS) (timeout of 60 seconds)' failed due to the following reason: The input cannot be shown compatible 2) 'WithProblem (timeout of 60 seconds)' failed due to the following reason: We add the following dependency tuples: Strict DPs: { pairNs^#() -> c_1(cons^#(0(), n__incr(oddNs())), oddNs^#()) , cons^#(X1, X2) -> c_2() , oddNs^#() -> c_3(incr^#(pairNs()), pairNs^#()) , incr^#(X) -> c_4() , activate^#(X) -> c_5() , activate^#(n__incr(X)) -> c_6(incr^#(X)) , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , activate^#(n__zip(X1, X2)) -> c_8(zip^#(X1, X2)) , activate^#(n__cons(X1, X2)) -> c_9(cons^#(X1, X2)) , activate^#(n__repItems(X)) -> c_10(repItems^#(X)) , take^#(X1, X2) -> c_11() , take^#(0(), XS) -> c_12() , zip^#(X1, X2) -> c_13() , zip^#(X, nil()) -> c_14() , zip^#(nil(), XS) -> c_15() , repItems^#(X) -> c_16() , repItems^#(nil()) -> c_17() } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { pairNs^#() -> c_1(cons^#(0(), n__incr(oddNs())), oddNs^#()) , cons^#(X1, X2) -> c_2() , oddNs^#() -> c_3(incr^#(pairNs()), pairNs^#()) , incr^#(X) -> c_4() , activate^#(X) -> c_5() , activate^#(n__incr(X)) -> c_6(incr^#(X)) , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , activate^#(n__zip(X1, X2)) -> c_8(zip^#(X1, X2)) , activate^#(n__cons(X1, X2)) -> c_9(cons^#(X1, X2)) , activate^#(n__repItems(X)) -> c_10(repItems^#(X)) , take^#(X1, X2) -> c_11() , take^#(0(), XS) -> c_12() , zip^#(X1, X2) -> c_13() , zip^#(X, nil()) -> c_14() , zip^#(nil(), XS) -> c_15() , repItems^#(X) -> c_16() , repItems^#(nil()) -> c_17() } Weak Trs: { pairNs() -> cons(0(), n__incr(oddNs())) , cons(X1, X2) -> n__cons(X1, X2) , oddNs() -> incr(pairNs()) , incr(X) -> n__incr(X) , activate(X) -> X , activate(n__incr(X)) -> incr(X) , activate(n__take(X1, X2)) -> take(X1, X2) , activate(n__zip(X1, X2)) -> zip(X1, X2) , activate(n__cons(X1, X2)) -> cons(X1, X2) , activate(n__repItems(X)) -> repItems(X) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , zip(X1, X2) -> n__zip(X1, X2) , zip(X, nil()) -> nil() , zip(nil(), XS) -> nil() , repItems(X) -> n__repItems(X) , repItems(nil()) -> nil() } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {2,4,5,11,12,13,14,15,16,17} by applications of Pre({2,4,5,11,12,13,14,15,16,17}) = {1,3,6,7,8,9,10}. Here rules are labeled as follows: DPs: { 1: pairNs^#() -> c_1(cons^#(0(), n__incr(oddNs())), oddNs^#()) , 2: cons^#(X1, X2) -> c_2() , 3: oddNs^#() -> c_3(incr^#(pairNs()), pairNs^#()) , 4: incr^#(X) -> c_4() , 5: activate^#(X) -> c_5() , 6: activate^#(n__incr(X)) -> c_6(incr^#(X)) , 7: activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , 8: activate^#(n__zip(X1, X2)) -> c_8(zip^#(X1, X2)) , 9: activate^#(n__cons(X1, X2)) -> c_9(cons^#(X1, X2)) , 10: activate^#(n__repItems(X)) -> c_10(repItems^#(X)) , 11: take^#(X1, X2) -> c_11() , 12: take^#(0(), XS) -> c_12() , 13: zip^#(X1, X2) -> c_13() , 14: zip^#(X, nil()) -> c_14() , 15: zip^#(nil(), XS) -> c_15() , 16: repItems^#(X) -> c_16() , 17: repItems^#(nil()) -> c_17() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { pairNs^#() -> c_1(cons^#(0(), n__incr(oddNs())), oddNs^#()) , oddNs^#() -> c_3(incr^#(pairNs()), pairNs^#()) , activate^#(n__incr(X)) -> c_6(incr^#(X)) , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , activate^#(n__zip(X1, X2)) -> c_8(zip^#(X1, X2)) , activate^#(n__cons(X1, X2)) -> c_9(cons^#(X1, X2)) , activate^#(n__repItems(X)) -> c_10(repItems^#(X)) } Weak DPs: { cons^#(X1, X2) -> c_2() , incr^#(X) -> c_4() , activate^#(X) -> c_5() , take^#(X1, X2) -> c_11() , take^#(0(), XS) -> c_12() , zip^#(X1, X2) -> c_13() , zip^#(X, nil()) -> c_14() , zip^#(nil(), XS) -> c_15() , repItems^#(X) -> c_16() , repItems^#(nil()) -> c_17() } Weak Trs: { pairNs() -> cons(0(), n__incr(oddNs())) , cons(X1, X2) -> n__cons(X1, X2) , oddNs() -> incr(pairNs()) , incr(X) -> n__incr(X) , activate(X) -> X , activate(n__incr(X)) -> incr(X) , activate(n__take(X1, X2)) -> take(X1, X2) , activate(n__zip(X1, X2)) -> zip(X1, X2) , activate(n__cons(X1, X2)) -> cons(X1, X2) , activate(n__repItems(X)) -> repItems(X) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , zip(X1, X2) -> n__zip(X1, X2) , zip(X, nil()) -> nil() , zip(nil(), XS) -> nil() , repItems(X) -> n__repItems(X) , repItems(nil()) -> nil() } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {3,4,5,6,7} by applications of Pre({3,4,5,6,7}) = {}. Here rules are labeled as follows: DPs: { 1: pairNs^#() -> c_1(cons^#(0(), n__incr(oddNs())), oddNs^#()) , 2: oddNs^#() -> c_3(incr^#(pairNs()), pairNs^#()) , 3: activate^#(n__incr(X)) -> c_6(incr^#(X)) , 4: activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , 5: activate^#(n__zip(X1, X2)) -> c_8(zip^#(X1, X2)) , 6: activate^#(n__cons(X1, X2)) -> c_9(cons^#(X1, X2)) , 7: activate^#(n__repItems(X)) -> c_10(repItems^#(X)) , 8: cons^#(X1, X2) -> c_2() , 9: incr^#(X) -> c_4() , 10: activate^#(X) -> c_5() , 11: take^#(X1, X2) -> c_11() , 12: take^#(0(), XS) -> c_12() , 13: zip^#(X1, X2) -> c_13() , 14: zip^#(X, nil()) -> c_14() , 15: zip^#(nil(), XS) -> c_15() , 16: repItems^#(X) -> c_16() , 17: repItems^#(nil()) -> c_17() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { pairNs^#() -> c_1(cons^#(0(), n__incr(oddNs())), oddNs^#()) , oddNs^#() -> c_3(incr^#(pairNs()), pairNs^#()) } Weak DPs: { cons^#(X1, X2) -> c_2() , incr^#(X) -> c_4() , activate^#(X) -> c_5() , activate^#(n__incr(X)) -> c_6(incr^#(X)) , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , activate^#(n__zip(X1, X2)) -> c_8(zip^#(X1, X2)) , activate^#(n__cons(X1, X2)) -> c_9(cons^#(X1, X2)) , activate^#(n__repItems(X)) -> c_10(repItems^#(X)) , take^#(X1, X2) -> c_11() , take^#(0(), XS) -> c_12() , zip^#(X1, X2) -> c_13() , zip^#(X, nil()) -> c_14() , zip^#(nil(), XS) -> c_15() , repItems^#(X) -> c_16() , repItems^#(nil()) -> c_17() } Weak Trs: { pairNs() -> cons(0(), n__incr(oddNs())) , cons(X1, X2) -> n__cons(X1, X2) , oddNs() -> incr(pairNs()) , incr(X) -> n__incr(X) , activate(X) -> X , activate(n__incr(X)) -> incr(X) , activate(n__take(X1, X2)) -> take(X1, X2) , activate(n__zip(X1, X2)) -> zip(X1, X2) , activate(n__cons(X1, X2)) -> cons(X1, X2) , activate(n__repItems(X)) -> repItems(X) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , zip(X1, X2) -> n__zip(X1, X2) , zip(X, nil()) -> nil() , zip(nil(), XS) -> nil() , repItems(X) -> n__repItems(X) , repItems(nil()) -> 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. { cons^#(X1, X2) -> c_2() , incr^#(X) -> c_4() , activate^#(X) -> c_5() , activate^#(n__incr(X)) -> c_6(incr^#(X)) , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , activate^#(n__zip(X1, X2)) -> c_8(zip^#(X1, X2)) , activate^#(n__cons(X1, X2)) -> c_9(cons^#(X1, X2)) , activate^#(n__repItems(X)) -> c_10(repItems^#(X)) , take^#(X1, X2) -> c_11() , take^#(0(), XS) -> c_12() , zip^#(X1, X2) -> c_13() , zip^#(X, nil()) -> c_14() , zip^#(nil(), XS) -> c_15() , repItems^#(X) -> c_16() , repItems^#(nil()) -> c_17() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { pairNs^#() -> c_1(cons^#(0(), n__incr(oddNs())), oddNs^#()) , oddNs^#() -> c_3(incr^#(pairNs()), pairNs^#()) } Weak Trs: { pairNs() -> cons(0(), n__incr(oddNs())) , cons(X1, X2) -> n__cons(X1, X2) , oddNs() -> incr(pairNs()) , incr(X) -> n__incr(X) , activate(X) -> X , activate(n__incr(X)) -> incr(X) , activate(n__take(X1, X2)) -> take(X1, X2) , activate(n__zip(X1, X2)) -> zip(X1, X2) , activate(n__cons(X1, X2)) -> cons(X1, X2) , activate(n__repItems(X)) -> repItems(X) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , zip(X1, X2) -> n__zip(X1, X2) , zip(X, nil()) -> nil() , zip(nil(), XS) -> nil() , repItems(X) -> n__repItems(X) , repItems(nil()) -> 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: { pairNs^#() -> c_1(cons^#(0(), n__incr(oddNs())), oddNs^#()) , oddNs^#() -> c_3(incr^#(pairNs()), pairNs^#()) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { pairNs^#() -> c_1(oddNs^#()) , oddNs^#() -> c_2(pairNs^#()) } Weak Trs: { pairNs() -> cons(0(), n__incr(oddNs())) , cons(X1, X2) -> n__cons(X1, X2) , oddNs() -> incr(pairNs()) , incr(X) -> n__incr(X) , activate(X) -> X , activate(n__incr(X)) -> incr(X) , activate(n__take(X1, X2)) -> take(X1, X2) , activate(n__zip(X1, X2)) -> zip(X1, X2) , activate(n__cons(X1, X2)) -> cons(X1, X2) , activate(n__repItems(X)) -> repItems(X) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , zip(X1, X2) -> n__zip(X1, X2) , zip(X, nil()) -> nil() , zip(nil(), XS) -> nil() , repItems(X) -> n__repItems(X) , repItems(nil()) -> nil() } Obligation: innermost runtime complexity Answer: MAYBE No rule is usable, rules are removed from the input problem. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { pairNs^#() -> c_1(oddNs^#()) , oddNs^#() -> c_2(pairNs^#()) } Obligation: innermost runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'WithProblem' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'Fastest' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'WithProblem' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'Polynomial Path Order (PS)' failed due to the following reason: The input cannot be shown compatible 2) 'Fastest (timeout of 5 seconds)' failed due to the following reason: Computation stopped due to timeout after 5.0 seconds. 3) 'Polynomial Path Order (PS)' failed due to the following reason: The input cannot be shown compatible Arrrr..