MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { a__pairNs() -> cons(0(), incr(oddNs())) , a__pairNs() -> pairNs() , a__oddNs() -> oddNs() , a__oddNs() -> a__incr(a__pairNs()) , a__incr(X) -> incr(X) , a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(0()) -> 0() , mark(incr(X)) -> a__incr(mark(X)) , mark(oddNs()) -> a__oddNs() , mark(s(X)) -> s(mark(X)) , mark(nil()) -> nil() , mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) , mark(pair(X1, X2)) -> pair(mark(X1), mark(X2)) , mark(zip(X1, X2)) -> a__zip(mark(X1), mark(X2)) , mark(repItems(X)) -> a__repItems(mark(X)) , mark(pairNs()) -> a__pairNs() , mark(tail(X)) -> a__tail(mark(X)) , a__take(X1, X2) -> take(X1, X2) , a__take(0(), XS) -> nil() , a__take(s(N), cons(X, XS)) -> cons(mark(X), take(N, XS)) , a__zip(X1, X2) -> zip(X1, X2) , a__zip(X, nil()) -> nil() , a__zip(cons(X, XS), cons(Y, YS)) -> cons(pair(mark(X), mark(Y)), zip(XS, YS)) , a__zip(nil(), XS) -> nil() , a__tail(X) -> tail(X) , a__tail(cons(X, XS)) -> mark(XS) , a__repItems(X) -> repItems(X) , a__repItems(cons(X, XS)) -> cons(mark(X), cons(X, repItems(XS))) , a__repItems(nil()) -> nil() } Obligation: runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'WithProblem (timeout of 60 seconds)' failed due to the following reason: Computation stopped due to timeout after 60.0 seconds. 2) '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: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'Bounds with perSymbol-enrichment and initial automaton 'match'' failed due to the following reason: match-boundness of the problem could not be verified. 2) 'Bounds with minimal-enrichment and initial automaton 'match'' failed due to the following reason: match-boundness of the problem could not be verified. 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 processor is inapplicable, reason: Processor only applicable for innermost runtime complexity analysis 2) 'Polynomial Path Order (PS) (timeout of 60 seconds)' failed due to the following reason: The processor is inapplicable, reason: Processor only applicable for innermost runtime complexity analysis 3) 'Innermost Weak Dependency Pairs (timeout of 60 seconds)' failed due to the following reason: We add the following weak dependency pairs: Strict DPs: { a__pairNs^#() -> c_1() , a__pairNs^#() -> c_2() , a__oddNs^#() -> c_3() , a__oddNs^#() -> c_4(a__incr^#(a__pairNs())) , a__incr^#(X) -> c_5(X) , a__incr^#(cons(X, XS)) -> c_6(mark^#(X), XS) , mark^#(cons(X1, X2)) -> c_7(mark^#(X1), X2) , mark^#(0()) -> c_8() , mark^#(incr(X)) -> c_9(a__incr^#(mark(X))) , mark^#(oddNs()) -> c_10(a__oddNs^#()) , mark^#(s(X)) -> c_11(mark^#(X)) , mark^#(nil()) -> c_12() , mark^#(take(X1, X2)) -> c_13(a__take^#(mark(X1), mark(X2))) , mark^#(pair(X1, X2)) -> c_14(mark^#(X1), mark^#(X2)) , mark^#(zip(X1, X2)) -> c_15(a__zip^#(mark(X1), mark(X2))) , mark^#(repItems(X)) -> c_16(a__repItems^#(mark(X))) , mark^#(pairNs()) -> c_17(a__pairNs^#()) , mark^#(tail(X)) -> c_18(a__tail^#(mark(X))) , a__take^#(X1, X2) -> c_19(X1, X2) , a__take^#(0(), XS) -> c_20() , a__take^#(s(N), cons(X, XS)) -> c_21(mark^#(X), N, XS) , a__zip^#(X1, X2) -> c_22(X1, X2) , a__zip^#(X, nil()) -> c_23() , a__zip^#(cons(X, XS), cons(Y, YS)) -> c_24(mark^#(X), mark^#(Y), XS, YS) , a__zip^#(nil(), XS) -> c_25() , a__repItems^#(X) -> c_28(X) , a__repItems^#(cons(X, XS)) -> c_29(mark^#(X), X, XS) , a__repItems^#(nil()) -> c_30() , a__tail^#(X) -> c_26(X) , a__tail^#(cons(X, XS)) -> c_27(mark^#(XS)) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__pairNs^#() -> c_1() , a__pairNs^#() -> c_2() , a__oddNs^#() -> c_3() , a__oddNs^#() -> c_4(a__incr^#(a__pairNs())) , a__incr^#(X) -> c_5(X) , a__incr^#(cons(X, XS)) -> c_6(mark^#(X), XS) , mark^#(cons(X1, X2)) -> c_7(mark^#(X1), X2) , mark^#(0()) -> c_8() , mark^#(incr(X)) -> c_9(a__incr^#(mark(X))) , mark^#(oddNs()) -> c_10(a__oddNs^#()) , mark^#(s(X)) -> c_11(mark^#(X)) , mark^#(nil()) -> c_12() , mark^#(take(X1, X2)) -> c_13(a__take^#(mark(X1), mark(X2))) , mark^#(pair(X1, X2)) -> c_14(mark^#(X1), mark^#(X2)) , mark^#(zip(X1, X2)) -> c_15(a__zip^#(mark(X1), mark(X2))) , mark^#(repItems(X)) -> c_16(a__repItems^#(mark(X))) , mark^#(pairNs()) -> c_17(a__pairNs^#()) , mark^#(tail(X)) -> c_18(a__tail^#(mark(X))) , a__take^#(X1, X2) -> c_19(X1, X2) , a__take^#(0(), XS) -> c_20() , a__take^#(s(N), cons(X, XS)) -> c_21(mark^#(X), N, XS) , a__zip^#(X1, X2) -> c_22(X1, X2) , a__zip^#(X, nil()) -> c_23() , a__zip^#(cons(X, XS), cons(Y, YS)) -> c_24(mark^#(X), mark^#(Y), XS, YS) , a__zip^#(nil(), XS) -> c_25() , a__repItems^#(X) -> c_28(X) , a__repItems^#(cons(X, XS)) -> c_29(mark^#(X), X, XS) , a__repItems^#(nil()) -> c_30() , a__tail^#(X) -> c_26(X) , a__tail^#(cons(X, XS)) -> c_27(mark^#(XS)) } Strict Trs: { a__pairNs() -> cons(0(), incr(oddNs())) , a__pairNs() -> pairNs() , a__oddNs() -> oddNs() , a__oddNs() -> a__incr(a__pairNs()) , a__incr(X) -> incr(X) , a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(0()) -> 0() , mark(incr(X)) -> a__incr(mark(X)) , mark(oddNs()) -> a__oddNs() , mark(s(X)) -> s(mark(X)) , mark(nil()) -> nil() , mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) , mark(pair(X1, X2)) -> pair(mark(X1), mark(X2)) , mark(zip(X1, X2)) -> a__zip(mark(X1), mark(X2)) , mark(repItems(X)) -> a__repItems(mark(X)) , mark(pairNs()) -> a__pairNs() , mark(tail(X)) -> a__tail(mark(X)) , a__take(X1, X2) -> take(X1, X2) , a__take(0(), XS) -> nil() , a__take(s(N), cons(X, XS)) -> cons(mark(X), take(N, XS)) , a__zip(X1, X2) -> zip(X1, X2) , a__zip(X, nil()) -> nil() , a__zip(cons(X, XS), cons(Y, YS)) -> cons(pair(mark(X), mark(Y)), zip(XS, YS)) , a__zip(nil(), XS) -> nil() , a__tail(X) -> tail(X) , a__tail(cons(X, XS)) -> mark(XS) , a__repItems(X) -> repItems(X) , a__repItems(cons(X, XS)) -> cons(mark(X), cons(X, repItems(XS))) , a__repItems(nil()) -> nil() } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {1,2,3,8,12,20,23,25,28} by applications of Pre({1,2,3,8,12,20,23,25,28}) = {5,6,7,10,11,13,14,15,16,17,19,21,22,24,26,27,29,30}. Here rules are labeled as follows: DPs: { 1: a__pairNs^#() -> c_1() , 2: a__pairNs^#() -> c_2() , 3: a__oddNs^#() -> c_3() , 4: a__oddNs^#() -> c_4(a__incr^#(a__pairNs())) , 5: a__incr^#(X) -> c_5(X) , 6: a__incr^#(cons(X, XS)) -> c_6(mark^#(X), XS) , 7: mark^#(cons(X1, X2)) -> c_7(mark^#(X1), X2) , 8: mark^#(0()) -> c_8() , 9: mark^#(incr(X)) -> c_9(a__incr^#(mark(X))) , 10: mark^#(oddNs()) -> c_10(a__oddNs^#()) , 11: mark^#(s(X)) -> c_11(mark^#(X)) , 12: mark^#(nil()) -> c_12() , 13: mark^#(take(X1, X2)) -> c_13(a__take^#(mark(X1), mark(X2))) , 14: mark^#(pair(X1, X2)) -> c_14(mark^#(X1), mark^#(X2)) , 15: mark^#(zip(X1, X2)) -> c_15(a__zip^#(mark(X1), mark(X2))) , 16: mark^#(repItems(X)) -> c_16(a__repItems^#(mark(X))) , 17: mark^#(pairNs()) -> c_17(a__pairNs^#()) , 18: mark^#(tail(X)) -> c_18(a__tail^#(mark(X))) , 19: a__take^#(X1, X2) -> c_19(X1, X2) , 20: a__take^#(0(), XS) -> c_20() , 21: a__take^#(s(N), cons(X, XS)) -> c_21(mark^#(X), N, XS) , 22: a__zip^#(X1, X2) -> c_22(X1, X2) , 23: a__zip^#(X, nil()) -> c_23() , 24: a__zip^#(cons(X, XS), cons(Y, YS)) -> c_24(mark^#(X), mark^#(Y), XS, YS) , 25: a__zip^#(nil(), XS) -> c_25() , 26: a__repItems^#(X) -> c_28(X) , 27: a__repItems^#(cons(X, XS)) -> c_29(mark^#(X), X, XS) , 28: a__repItems^#(nil()) -> c_30() , 29: a__tail^#(X) -> c_26(X) , 30: a__tail^#(cons(X, XS)) -> c_27(mark^#(XS)) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__oddNs^#() -> c_4(a__incr^#(a__pairNs())) , a__incr^#(X) -> c_5(X) , a__incr^#(cons(X, XS)) -> c_6(mark^#(X), XS) , mark^#(cons(X1, X2)) -> c_7(mark^#(X1), X2) , mark^#(incr(X)) -> c_9(a__incr^#(mark(X))) , mark^#(oddNs()) -> c_10(a__oddNs^#()) , mark^#(s(X)) -> c_11(mark^#(X)) , mark^#(take(X1, X2)) -> c_13(a__take^#(mark(X1), mark(X2))) , mark^#(pair(X1, X2)) -> c_14(mark^#(X1), mark^#(X2)) , mark^#(zip(X1, X2)) -> c_15(a__zip^#(mark(X1), mark(X2))) , mark^#(repItems(X)) -> c_16(a__repItems^#(mark(X))) , mark^#(pairNs()) -> c_17(a__pairNs^#()) , mark^#(tail(X)) -> c_18(a__tail^#(mark(X))) , a__take^#(X1, X2) -> c_19(X1, X2) , a__take^#(s(N), cons(X, XS)) -> c_21(mark^#(X), N, XS) , a__zip^#(X1, X2) -> c_22(X1, X2) , a__zip^#(cons(X, XS), cons(Y, YS)) -> c_24(mark^#(X), mark^#(Y), XS, YS) , a__repItems^#(X) -> c_28(X) , a__repItems^#(cons(X, XS)) -> c_29(mark^#(X), X, XS) , a__tail^#(X) -> c_26(X) , a__tail^#(cons(X, XS)) -> c_27(mark^#(XS)) } Strict Trs: { a__pairNs() -> cons(0(), incr(oddNs())) , a__pairNs() -> pairNs() , a__oddNs() -> oddNs() , a__oddNs() -> a__incr(a__pairNs()) , a__incr(X) -> incr(X) , a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(0()) -> 0() , mark(incr(X)) -> a__incr(mark(X)) , mark(oddNs()) -> a__oddNs() , mark(s(X)) -> s(mark(X)) , mark(nil()) -> nil() , mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) , mark(pair(X1, X2)) -> pair(mark(X1), mark(X2)) , mark(zip(X1, X2)) -> a__zip(mark(X1), mark(X2)) , mark(repItems(X)) -> a__repItems(mark(X)) , mark(pairNs()) -> a__pairNs() , mark(tail(X)) -> a__tail(mark(X)) , a__take(X1, X2) -> take(X1, X2) , a__take(0(), XS) -> nil() , a__take(s(N), cons(X, XS)) -> cons(mark(X), take(N, XS)) , a__zip(X1, X2) -> zip(X1, X2) , a__zip(X, nil()) -> nil() , a__zip(cons(X, XS), cons(Y, YS)) -> cons(pair(mark(X), mark(Y)), zip(XS, YS)) , a__zip(nil(), XS) -> nil() , a__tail(X) -> tail(X) , a__tail(cons(X, XS)) -> mark(XS) , a__repItems(X) -> repItems(X) , a__repItems(cons(X, XS)) -> cons(mark(X), cons(X, repItems(XS))) , a__repItems(nil()) -> nil() } Weak DPs: { a__pairNs^#() -> c_1() , a__pairNs^#() -> c_2() , a__oddNs^#() -> c_3() , mark^#(0()) -> c_8() , mark^#(nil()) -> c_12() , a__take^#(0(), XS) -> c_20() , a__zip^#(X, nil()) -> c_23() , a__zip^#(nil(), XS) -> c_25() , a__repItems^#(nil()) -> c_30() } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {12} by applications of Pre({12}) = {2,3,4,7,9,14,15,16,17,18,19,20,21}. Here rules are labeled as follows: DPs: { 1: a__oddNs^#() -> c_4(a__incr^#(a__pairNs())) , 2: a__incr^#(X) -> c_5(X) , 3: a__incr^#(cons(X, XS)) -> c_6(mark^#(X), XS) , 4: mark^#(cons(X1, X2)) -> c_7(mark^#(X1), X2) , 5: mark^#(incr(X)) -> c_9(a__incr^#(mark(X))) , 6: mark^#(oddNs()) -> c_10(a__oddNs^#()) , 7: mark^#(s(X)) -> c_11(mark^#(X)) , 8: mark^#(take(X1, X2)) -> c_13(a__take^#(mark(X1), mark(X2))) , 9: mark^#(pair(X1, X2)) -> c_14(mark^#(X1), mark^#(X2)) , 10: mark^#(zip(X1, X2)) -> c_15(a__zip^#(mark(X1), mark(X2))) , 11: mark^#(repItems(X)) -> c_16(a__repItems^#(mark(X))) , 12: mark^#(pairNs()) -> c_17(a__pairNs^#()) , 13: mark^#(tail(X)) -> c_18(a__tail^#(mark(X))) , 14: a__take^#(X1, X2) -> c_19(X1, X2) , 15: a__take^#(s(N), cons(X, XS)) -> c_21(mark^#(X), N, XS) , 16: a__zip^#(X1, X2) -> c_22(X1, X2) , 17: a__zip^#(cons(X, XS), cons(Y, YS)) -> c_24(mark^#(X), mark^#(Y), XS, YS) , 18: a__repItems^#(X) -> c_28(X) , 19: a__repItems^#(cons(X, XS)) -> c_29(mark^#(X), X, XS) , 20: a__tail^#(X) -> c_26(X) , 21: a__tail^#(cons(X, XS)) -> c_27(mark^#(XS)) , 22: a__pairNs^#() -> c_1() , 23: a__pairNs^#() -> c_2() , 24: a__oddNs^#() -> c_3() , 25: mark^#(0()) -> c_8() , 26: mark^#(nil()) -> c_12() , 27: a__take^#(0(), XS) -> c_20() , 28: a__zip^#(X, nil()) -> c_23() , 29: a__zip^#(nil(), XS) -> c_25() , 30: a__repItems^#(nil()) -> c_30() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__oddNs^#() -> c_4(a__incr^#(a__pairNs())) , a__incr^#(X) -> c_5(X) , a__incr^#(cons(X, XS)) -> c_6(mark^#(X), XS) , mark^#(cons(X1, X2)) -> c_7(mark^#(X1), X2) , mark^#(incr(X)) -> c_9(a__incr^#(mark(X))) , mark^#(oddNs()) -> c_10(a__oddNs^#()) , mark^#(s(X)) -> c_11(mark^#(X)) , mark^#(take(X1, X2)) -> c_13(a__take^#(mark(X1), mark(X2))) , mark^#(pair(X1, X2)) -> c_14(mark^#(X1), mark^#(X2)) , mark^#(zip(X1, X2)) -> c_15(a__zip^#(mark(X1), mark(X2))) , mark^#(repItems(X)) -> c_16(a__repItems^#(mark(X))) , mark^#(tail(X)) -> c_18(a__tail^#(mark(X))) , a__take^#(X1, X2) -> c_19(X1, X2) , a__take^#(s(N), cons(X, XS)) -> c_21(mark^#(X), N, XS) , a__zip^#(X1, X2) -> c_22(X1, X2) , a__zip^#(cons(X, XS), cons(Y, YS)) -> c_24(mark^#(X), mark^#(Y), XS, YS) , a__repItems^#(X) -> c_28(X) , a__repItems^#(cons(X, XS)) -> c_29(mark^#(X), X, XS) , a__tail^#(X) -> c_26(X) , a__tail^#(cons(X, XS)) -> c_27(mark^#(XS)) } Strict Trs: { a__pairNs() -> cons(0(), incr(oddNs())) , a__pairNs() -> pairNs() , a__oddNs() -> oddNs() , a__oddNs() -> a__incr(a__pairNs()) , a__incr(X) -> incr(X) , a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(0()) -> 0() , mark(incr(X)) -> a__incr(mark(X)) , mark(oddNs()) -> a__oddNs() , mark(s(X)) -> s(mark(X)) , mark(nil()) -> nil() , mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) , mark(pair(X1, X2)) -> pair(mark(X1), mark(X2)) , mark(zip(X1, X2)) -> a__zip(mark(X1), mark(X2)) , mark(repItems(X)) -> a__repItems(mark(X)) , mark(pairNs()) -> a__pairNs() , mark(tail(X)) -> a__tail(mark(X)) , a__take(X1, X2) -> take(X1, X2) , a__take(0(), XS) -> nil() , a__take(s(N), cons(X, XS)) -> cons(mark(X), take(N, XS)) , a__zip(X1, X2) -> zip(X1, X2) , a__zip(X, nil()) -> nil() , a__zip(cons(X, XS), cons(Y, YS)) -> cons(pair(mark(X), mark(Y)), zip(XS, YS)) , a__zip(nil(), XS) -> nil() , a__tail(X) -> tail(X) , a__tail(cons(X, XS)) -> mark(XS) , a__repItems(X) -> repItems(X) , a__repItems(cons(X, XS)) -> cons(mark(X), cons(X, repItems(XS))) , a__repItems(nil()) -> nil() } Weak DPs: { a__pairNs^#() -> c_1() , a__pairNs^#() -> c_2() , a__oddNs^#() -> c_3() , mark^#(0()) -> c_8() , mark^#(nil()) -> c_12() , mark^#(pairNs()) -> c_17(a__pairNs^#()) , a__take^#(0(), XS) -> c_20() , a__zip^#(X, nil()) -> c_23() , a__zip^#(nil(), XS) -> c_25() , a__repItems^#(nil()) -> c_30() } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. Arrrr..