MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { a__nats() -> a__adx(a__zeros()) , a__nats() -> nats() , a__adx(X) -> adx(X) , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) , a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__incr(X) -> incr(X) , a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) , a__hd(X) -> hd(X) , a__hd(cons(X, Y)) -> mark(X) , mark(cons(X1, X2)) -> cons(X1, X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(s(X)) -> s(X) , mark(incr(X)) -> a__incr(mark(X)) , mark(adx(X)) -> a__adx(mark(X)) , mark(nats()) -> a__nats() , mark(hd(X)) -> a__hd(mark(X)) , mark(tl(X)) -> a__tl(mark(X)) , a__tl(X) -> tl(X) , a__tl(cons(X, Y)) -> mark(Y) } Obligation: innermost runtime complexity Answer: MAYBE We add following dependency tuples: Strict DPs: { a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#()) , a__nats^#() -> c_2() , a__adx^#(X) -> c_3() , a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y)))) , a__zeros^#() -> c_5() , a__zeros^#() -> c_6() , a__incr^#(X) -> c_7() , a__incr^#(cons(X, Y)) -> c_8() , a__hd^#(X) -> c_9() , a__hd^#(cons(X, Y)) -> c_10(mark^#(X)) , mark^#(cons(X1, X2)) -> c_11() , mark^#(0()) -> c_12() , mark^#(zeros()) -> c_13(a__zeros^#()) , mark^#(s(X)) -> c_14() , mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X)) , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X)) , mark^#(nats()) -> c_17(a__nats^#()) , mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X)) , mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X)) , a__tl^#(X) -> c_20() , a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#()) , a__nats^#() -> c_2() , a__adx^#(X) -> c_3() , a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y)))) , a__zeros^#() -> c_5() , a__zeros^#() -> c_6() , a__incr^#(X) -> c_7() , a__incr^#(cons(X, Y)) -> c_8() , a__hd^#(X) -> c_9() , a__hd^#(cons(X, Y)) -> c_10(mark^#(X)) , mark^#(cons(X1, X2)) -> c_11() , mark^#(0()) -> c_12() , mark^#(zeros()) -> c_13(a__zeros^#()) , mark^#(s(X)) -> c_14() , mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X)) , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X)) , mark^#(nats()) -> c_17(a__nats^#()) , mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X)) , mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X)) , a__tl^#(X) -> c_20() , a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) } Weak Trs: { a__nats() -> a__adx(a__zeros()) , a__nats() -> nats() , a__adx(X) -> adx(X) , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) , a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__incr(X) -> incr(X) , a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) , a__hd(X) -> hd(X) , a__hd(cons(X, Y)) -> mark(X) , mark(cons(X1, X2)) -> cons(X1, X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(s(X)) -> s(X) , mark(incr(X)) -> a__incr(mark(X)) , mark(adx(X)) -> a__adx(mark(X)) , mark(nats()) -> a__nats() , mark(hd(X)) -> a__hd(mark(X)) , mark(tl(X)) -> a__tl(mark(X)) , a__tl(X) -> tl(X) , a__tl(cons(X, Y)) -> mark(Y) } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {2,3,5,6,7,8,9,11,12,14,20} by applications of Pre({2,3,5,6,7,8,9,11,12,14,20}) = {1,4,10,13,15,16,17,18,19,21}. Here rules are labeled as follows: DPs: { 1: a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#()) , 2: a__nats^#() -> c_2() , 3: a__adx^#(X) -> c_3() , 4: a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y)))) , 5: a__zeros^#() -> c_5() , 6: a__zeros^#() -> c_6() , 7: a__incr^#(X) -> c_7() , 8: a__incr^#(cons(X, Y)) -> c_8() , 9: a__hd^#(X) -> c_9() , 10: a__hd^#(cons(X, Y)) -> c_10(mark^#(X)) , 11: mark^#(cons(X1, X2)) -> c_11() , 12: mark^#(0()) -> c_12() , 13: mark^#(zeros()) -> c_13(a__zeros^#()) , 14: mark^#(s(X)) -> c_14() , 15: mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X)) , 16: mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X)) , 17: mark^#(nats()) -> c_17(a__nats^#()) , 18: mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X)) , 19: mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X)) , 20: a__tl^#(X) -> c_20() , 21: a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#()) , a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y)))) , a__hd^#(cons(X, Y)) -> c_10(mark^#(X)) , mark^#(zeros()) -> c_13(a__zeros^#()) , mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X)) , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X)) , mark^#(nats()) -> c_17(a__nats^#()) , mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X)) , mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X)) , a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) } Weak DPs: { a__nats^#() -> c_2() , a__adx^#(X) -> c_3() , a__zeros^#() -> c_5() , a__zeros^#() -> c_6() , a__incr^#(X) -> c_7() , a__incr^#(cons(X, Y)) -> c_8() , a__hd^#(X) -> c_9() , mark^#(cons(X1, X2)) -> c_11() , mark^#(0()) -> c_12() , mark^#(s(X)) -> c_14() , a__tl^#(X) -> c_20() } Weak Trs: { a__nats() -> a__adx(a__zeros()) , a__nats() -> nats() , a__adx(X) -> adx(X) , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) , a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__incr(X) -> incr(X) , a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) , a__hd(X) -> hd(X) , a__hd(cons(X, Y)) -> mark(X) , mark(cons(X1, X2)) -> cons(X1, X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(s(X)) -> s(X) , mark(incr(X)) -> a__incr(mark(X)) , mark(adx(X)) -> a__adx(mark(X)) , mark(nats()) -> a__nats() , mark(hd(X)) -> a__hd(mark(X)) , mark(tl(X)) -> a__tl(mark(X)) , a__tl(X) -> tl(X) , a__tl(cons(X, Y)) -> mark(Y) } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {2,4} by applications of Pre({2,4}) = {1,3,5,6,8,9,10}. Here rules are labeled as follows: DPs: { 1: a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#()) , 2: a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y)))) , 3: a__hd^#(cons(X, Y)) -> c_10(mark^#(X)) , 4: mark^#(zeros()) -> c_13(a__zeros^#()) , 5: mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X)) , 6: mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X)) , 7: mark^#(nats()) -> c_17(a__nats^#()) , 8: mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X)) , 9: mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X)) , 10: a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) , 11: a__nats^#() -> c_2() , 12: a__adx^#(X) -> c_3() , 13: a__zeros^#() -> c_5() , 14: a__zeros^#() -> c_6() , 15: a__incr^#(X) -> c_7() , 16: a__incr^#(cons(X, Y)) -> c_8() , 17: a__hd^#(X) -> c_9() , 18: mark^#(cons(X1, X2)) -> c_11() , 19: mark^#(0()) -> c_12() , 20: mark^#(s(X)) -> c_14() , 21: a__tl^#(X) -> c_20() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#()) , a__hd^#(cons(X, Y)) -> c_10(mark^#(X)) , mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X)) , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X)) , mark^#(nats()) -> c_17(a__nats^#()) , mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X)) , mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X)) , a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) } Weak DPs: { a__nats^#() -> c_2() , a__adx^#(X) -> c_3() , a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y)))) , a__zeros^#() -> c_5() , a__zeros^#() -> c_6() , a__incr^#(X) -> c_7() , a__incr^#(cons(X, Y)) -> c_8() , a__hd^#(X) -> c_9() , mark^#(cons(X1, X2)) -> c_11() , mark^#(0()) -> c_12() , mark^#(zeros()) -> c_13(a__zeros^#()) , mark^#(s(X)) -> c_14() , a__tl^#(X) -> c_20() } Weak Trs: { a__nats() -> a__adx(a__zeros()) , a__nats() -> nats() , a__adx(X) -> adx(X) , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) , a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__incr(X) -> incr(X) , a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) , a__hd(X) -> hd(X) , a__hd(cons(X, Y)) -> mark(X) , mark(cons(X1, X2)) -> cons(X1, X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(s(X)) -> s(X) , mark(incr(X)) -> a__incr(mark(X)) , mark(adx(X)) -> a__adx(mark(X)) , mark(nats()) -> a__nats() , mark(hd(X)) -> a__hd(mark(X)) , mark(tl(X)) -> a__tl(mark(X)) , a__tl(X) -> tl(X) , a__tl(cons(X, Y)) -> mark(Y) } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {1} by applications of Pre({1}) = {5}. Here rules are labeled as follows: DPs: { 1: a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#()) , 2: a__hd^#(cons(X, Y)) -> c_10(mark^#(X)) , 3: mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X)) , 4: mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X)) , 5: mark^#(nats()) -> c_17(a__nats^#()) , 6: mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X)) , 7: mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X)) , 8: a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) , 9: a__nats^#() -> c_2() , 10: a__adx^#(X) -> c_3() , 11: a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y)))) , 12: a__zeros^#() -> c_5() , 13: a__zeros^#() -> c_6() , 14: a__incr^#(X) -> c_7() , 15: a__incr^#(cons(X, Y)) -> c_8() , 16: a__hd^#(X) -> c_9() , 17: mark^#(cons(X1, X2)) -> c_11() , 18: mark^#(0()) -> c_12() , 19: mark^#(zeros()) -> c_13(a__zeros^#()) , 20: mark^#(s(X)) -> c_14() , 21: a__tl^#(X) -> c_20() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__hd^#(cons(X, Y)) -> c_10(mark^#(X)) , mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X)) , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X)) , mark^#(nats()) -> c_17(a__nats^#()) , mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X)) , mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X)) , a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) } Weak DPs: { a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#()) , a__nats^#() -> c_2() , a__adx^#(X) -> c_3() , a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y)))) , a__zeros^#() -> c_5() , a__zeros^#() -> c_6() , a__incr^#(X) -> c_7() , a__incr^#(cons(X, Y)) -> c_8() , a__hd^#(X) -> c_9() , mark^#(cons(X1, X2)) -> c_11() , mark^#(0()) -> c_12() , mark^#(zeros()) -> c_13(a__zeros^#()) , mark^#(s(X)) -> c_14() , a__tl^#(X) -> c_20() } Weak Trs: { a__nats() -> a__adx(a__zeros()) , a__nats() -> nats() , a__adx(X) -> adx(X) , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) , a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__incr(X) -> incr(X) , a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) , a__hd(X) -> hd(X) , a__hd(cons(X, Y)) -> mark(X) , mark(cons(X1, X2)) -> cons(X1, X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(s(X)) -> s(X) , mark(incr(X)) -> a__incr(mark(X)) , mark(adx(X)) -> a__adx(mark(X)) , mark(nats()) -> a__nats() , mark(hd(X)) -> a__hd(mark(X)) , mark(tl(X)) -> a__tl(mark(X)) , a__tl(X) -> tl(X) , a__tl(cons(X, Y)) -> mark(Y) } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {4} by applications of Pre({4}) = {1,2,3,5,6,7}. Here rules are labeled as follows: DPs: { 1: a__hd^#(cons(X, Y)) -> c_10(mark^#(X)) , 2: mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X)) , 3: mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X)) , 4: mark^#(nats()) -> c_17(a__nats^#()) , 5: mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X)) , 6: mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X)) , 7: a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) , 8: a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#()) , 9: a__nats^#() -> c_2() , 10: a__adx^#(X) -> c_3() , 11: a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y)))) , 12: a__zeros^#() -> c_5() , 13: a__zeros^#() -> c_6() , 14: a__incr^#(X) -> c_7() , 15: a__incr^#(cons(X, Y)) -> c_8() , 16: a__hd^#(X) -> c_9() , 17: mark^#(cons(X1, X2)) -> c_11() , 18: mark^#(0()) -> c_12() , 19: mark^#(zeros()) -> c_13(a__zeros^#()) , 20: mark^#(s(X)) -> c_14() , 21: a__tl^#(X) -> c_20() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__hd^#(cons(X, Y)) -> c_10(mark^#(X)) , mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X)) , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X)) , mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X)) , mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X)) , a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) } Weak DPs: { a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#()) , a__nats^#() -> c_2() , a__adx^#(X) -> c_3() , a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y)))) , a__zeros^#() -> c_5() , a__zeros^#() -> c_6() , a__incr^#(X) -> c_7() , a__incr^#(cons(X, Y)) -> c_8() , a__hd^#(X) -> c_9() , mark^#(cons(X1, X2)) -> c_11() , mark^#(0()) -> c_12() , mark^#(zeros()) -> c_13(a__zeros^#()) , mark^#(s(X)) -> c_14() , mark^#(nats()) -> c_17(a__nats^#()) , a__tl^#(X) -> c_20() } Weak Trs: { a__nats() -> a__adx(a__zeros()) , a__nats() -> nats() , a__adx(X) -> adx(X) , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) , a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__incr(X) -> incr(X) , a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) , a__hd(X) -> hd(X) , a__hd(cons(X, Y)) -> mark(X) , mark(cons(X1, X2)) -> cons(X1, X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(s(X)) -> s(X) , mark(incr(X)) -> a__incr(mark(X)) , mark(adx(X)) -> a__adx(mark(X)) , mark(nats()) -> a__nats() , mark(hd(X)) -> a__hd(mark(X)) , mark(tl(X)) -> a__tl(mark(X)) , a__tl(X) -> tl(X) , a__tl(cons(X, Y)) -> mark(Y) } 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__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#()) , a__nats^#() -> c_2() , a__adx^#(X) -> c_3() , a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y)))) , a__zeros^#() -> c_5() , a__zeros^#() -> c_6() , a__incr^#(X) -> c_7() , a__incr^#(cons(X, Y)) -> c_8() , a__hd^#(X) -> c_9() , mark^#(cons(X1, X2)) -> c_11() , mark^#(0()) -> c_12() , mark^#(zeros()) -> c_13(a__zeros^#()) , mark^#(s(X)) -> c_14() , mark^#(nats()) -> c_17(a__nats^#()) , a__tl^#(X) -> c_20() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__hd^#(cons(X, Y)) -> c_10(mark^#(X)) , mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X)) , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X)) , mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X)) , mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X)) , a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) } Weak Trs: { a__nats() -> a__adx(a__zeros()) , a__nats() -> nats() , a__adx(X) -> adx(X) , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) , a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__incr(X) -> incr(X) , a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) , a__hd(X) -> hd(X) , a__hd(cons(X, Y)) -> mark(X) , mark(cons(X1, X2)) -> cons(X1, X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(s(X)) -> s(X) , mark(incr(X)) -> a__incr(mark(X)) , mark(adx(X)) -> a__adx(mark(X)) , mark(nats()) -> a__nats() , mark(hd(X)) -> a__hd(mark(X)) , mark(tl(X)) -> a__tl(mark(X)) , a__tl(X) -> tl(X) , a__tl(cons(X, Y)) -> mark(Y) } Obligation: innermost runtime complexity Answer: MAYBE Due to missing edges in the dependency-graph, the right-hand sides of following rules could be simplified: { mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X)) , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X)) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__hd^#(cons(X, Y)) -> c_1(mark^#(X)) , mark^#(incr(X)) -> c_2(mark^#(X)) , mark^#(adx(X)) -> c_3(mark^#(X)) , mark^#(hd(X)) -> c_4(a__hd^#(mark(X)), mark^#(X)) , mark^#(tl(X)) -> c_5(a__tl^#(mark(X)), mark^#(X)) , a__tl^#(cons(X, Y)) -> c_6(mark^#(Y)) } Weak Trs: { a__nats() -> a__adx(a__zeros()) , a__nats() -> nats() , a__adx(X) -> adx(X) , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) , a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__incr(X) -> incr(X) , a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) , a__hd(X) -> hd(X) , a__hd(cons(X, Y)) -> mark(X) , mark(cons(X1, X2)) -> cons(X1, X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(s(X)) -> s(X) , mark(incr(X)) -> a__incr(mark(X)) , mark(adx(X)) -> a__adx(mark(X)) , mark(nats()) -> a__nats() , mark(hd(X)) -> a__hd(mark(X)) , mark(tl(X)) -> a__tl(mark(X)) , a__tl(X) -> tl(X) , a__tl(cons(X, Y)) -> mark(Y) } Obligation: innermost runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'matrices' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'matrix interpretation of dimension 4' failed due to the following reason: The input cannot be shown compatible 2) 'matrix interpretation of dimension 3' failed due to the following reason: The input cannot be shown compatible 3) 'matrix interpretation of dimension 3' failed due to the following reason: The input cannot be shown compatible 4) 'matrix interpretation of dimension 2' failed due to the following reason: The input cannot be shown compatible 5) 'matrix interpretation of dimension 2' failed due to the following reason: The input cannot be shown compatible 6) 'matrix interpretation of dimension 1' failed due to the following reason: The input cannot be shown compatible 2) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. Arrrr..