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(a__add(a__sqr(mark(X)), a__dbl(mark(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(mark(X)) , mark(0()) -> 0() , mark(nil()) -> nil() , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2)) , mark(dbl(X)) -> a__dbl(mark(X)) , mark(sqr(X)) -> a__sqr(mark(X)) , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2)) , mark(half(X)) -> a__half(mark(X)) , a__add(X1, X2) -> add(X1, X2) , a__add(s(X), Y) -> s(a__add(mark(X), mark(Y))) , a__add(0(), X) -> mark(X) , a__dbl(X) -> dbl(X) , a__dbl(s(X)) -> s(s(a__dbl(mark(X)))) , a__dbl(0()) -> 0() , 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() , a__half(X) -> half(X) , a__half(s(s(X))) -> s(a__half(mark(X))) , a__half(s(0())) -> 0() , a__half(0()) -> 0() , a__half(dbl(X)) -> mark(X) } 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__add^#(a__sqr(mark(X)), a__dbl(mark(X))), a__sqr^#(mark(X)), mark^#(X), a__dbl^#(mark(X)), mark^#(X)) , 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^#(X)) , mark^#(0()) -> c_10() , mark^#(nil()) -> c_11() , mark^#(first(X1, X2)) -> c_12(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X)), mark^#(X)) , mark^#(sqr(X)) -> c_14(a__sqr^#(mark(X)), mark^#(X)) , mark^#(add(X1, X2)) -> c_15(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(half(X)) -> c_16(a__half^#(mark(X)), mark^#(X)) , a__add^#(X1, X2) -> c_17() , a__add^#(s(X), Y) -> c_18(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , a__add^#(0(), X) -> c_19(mark^#(X)) , a__dbl^#(X) -> c_20() , a__dbl^#(s(X)) -> c_21(a__dbl^#(mark(X)), mark^#(X)) , a__dbl^#(0()) -> c_22() , a__first^#(X1, X2) -> c_23() , a__first^#(s(X), cons(Y, Z)) -> c_24(mark^#(Y)) , a__first^#(0(), X) -> c_25() , a__half^#(X) -> c_26() , a__half^#(s(s(X))) -> c_27(a__half^#(mark(X)), mark^#(X)) , a__half^#(s(0())) -> c_28() , a__half^#(0()) -> c_29() , a__half^#(dbl(X)) -> c_30(mark^#(X)) } 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__add^#(a__sqr(mark(X)), a__dbl(mark(X))), a__sqr^#(mark(X)), mark^#(X), a__dbl^#(mark(X)), mark^#(X)) , 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^#(X)) , mark^#(0()) -> c_10() , mark^#(nil()) -> c_11() , mark^#(first(X1, X2)) -> c_12(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X)), mark^#(X)) , mark^#(sqr(X)) -> c_14(a__sqr^#(mark(X)), mark^#(X)) , mark^#(add(X1, X2)) -> c_15(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(half(X)) -> c_16(a__half^#(mark(X)), mark^#(X)) , a__add^#(X1, X2) -> c_17() , a__add^#(s(X), Y) -> c_18(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , a__add^#(0(), X) -> c_19(mark^#(X)) , a__dbl^#(X) -> c_20() , a__dbl^#(s(X)) -> c_21(a__dbl^#(mark(X)), mark^#(X)) , a__dbl^#(0()) -> c_22() , a__first^#(X1, X2) -> c_23() , a__first^#(s(X), cons(Y, Z)) -> c_24(mark^#(Y)) , a__first^#(0(), X) -> c_25() , a__half^#(X) -> c_26() , a__half^#(s(s(X))) -> c_27(a__half^#(mark(X)), mark^#(X)) , a__half^#(s(0())) -> c_28() , a__half^#(0()) -> c_29() , a__half^#(dbl(X)) -> c_30(mark^#(X)) } 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(a__add(a__sqr(mark(X)), a__dbl(mark(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(mark(X)) , mark(0()) -> 0() , mark(nil()) -> nil() , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2)) , mark(dbl(X)) -> a__dbl(mark(X)) , mark(sqr(X)) -> a__sqr(mark(X)) , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2)) , mark(half(X)) -> a__half(mark(X)) , a__add(X1, X2) -> add(X1, X2) , a__add(s(X), Y) -> s(a__add(mark(X), mark(Y))) , a__add(0(), X) -> mark(X) , a__dbl(X) -> dbl(X) , a__dbl(s(X)) -> s(s(a__dbl(mark(X)))) , a__dbl(0()) -> 0() , 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() , a__half(X) -> half(X) , a__half(s(s(X))) -> s(a__half(mark(X))) , a__half(s(0())) -> 0() , a__half(0()) -> 0() , a__half(dbl(X)) -> mark(X) } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {2,3,5,10,11,17,20,22,23,25,26,28,29} by applications of Pre({2,3,5,10,11,17,20,22,23,25,26,28,29}) = {1,4,6,7,8,9,12,13,14,15,16,18,19,21,24,27,30}. 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(a__add^#(a__sqr(mark(X)), a__dbl(mark(X))), a__sqr^#(mark(X)), mark^#(X), a__dbl^#(mark(X)), mark^#(X)) , 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(mark^#(X)) , 10: mark^#(0()) -> c_10() , 11: mark^#(nil()) -> c_11() , 12: mark^#(first(X1, X2)) -> c_12(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , 13: mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X)), mark^#(X)) , 14: mark^#(sqr(X)) -> c_14(a__sqr^#(mark(X)), mark^#(X)) , 15: mark^#(add(X1, X2)) -> c_15(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , 16: mark^#(half(X)) -> c_16(a__half^#(mark(X)), mark^#(X)) , 17: a__add^#(X1, X2) -> c_17() , 18: a__add^#(s(X), Y) -> c_18(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , 19: a__add^#(0(), X) -> c_19(mark^#(X)) , 20: a__dbl^#(X) -> c_20() , 21: a__dbl^#(s(X)) -> c_21(a__dbl^#(mark(X)), mark^#(X)) , 22: a__dbl^#(0()) -> c_22() , 23: a__first^#(X1, X2) -> c_23() , 24: a__first^#(s(X), cons(Y, Z)) -> c_24(mark^#(Y)) , 25: a__first^#(0(), X) -> c_25() , 26: a__half^#(X) -> c_26() , 27: a__half^#(s(s(X))) -> c_27(a__half^#(mark(X)), mark^#(X)) , 28: a__half^#(s(0())) -> c_28() , 29: a__half^#(0()) -> c_29() , 30: a__half^#(dbl(X)) -> c_30(mark^#(X)) } 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__sqr^#(s(X)) -> c_4(a__add^#(a__sqr(mark(X)), a__dbl(mark(X))), a__sqr^#(mark(X)), mark^#(X), a__dbl^#(mark(X)), mark^#(X)) , 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^#(X)) , mark^#(first(X1, X2)) -> c_12(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X)), mark^#(X)) , mark^#(sqr(X)) -> c_14(a__sqr^#(mark(X)), mark^#(X)) , mark^#(add(X1, X2)) -> c_15(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(half(X)) -> c_16(a__half^#(mark(X)), mark^#(X)) , a__add^#(s(X), Y) -> c_18(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , a__add^#(0(), X) -> c_19(mark^#(X)) , a__dbl^#(s(X)) -> c_21(a__dbl^#(mark(X)), mark^#(X)) , a__first^#(s(X), cons(Y, Z)) -> c_24(mark^#(Y)) , a__half^#(s(s(X))) -> c_27(a__half^#(mark(X)), mark^#(X)) , a__half^#(dbl(X)) -> c_30(mark^#(X)) } Weak DPs: { a__terms^#(X) -> c_2() , a__sqr^#(X) -> c_3() , a__sqr^#(0()) -> c_5() , mark^#(0()) -> c_10() , mark^#(nil()) -> c_11() , a__add^#(X1, X2) -> c_17() , a__dbl^#(X) -> c_20() , a__dbl^#(0()) -> c_22() , a__first^#(X1, X2) -> c_23() , a__first^#(0(), X) -> c_25() , a__half^#(X) -> c_26() , a__half^#(s(0())) -> c_28() , a__half^#(0()) -> c_29() } 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(a__add(a__sqr(mark(X)), a__dbl(mark(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(mark(X)) , mark(0()) -> 0() , mark(nil()) -> nil() , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2)) , mark(dbl(X)) -> a__dbl(mark(X)) , mark(sqr(X)) -> a__sqr(mark(X)) , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2)) , mark(half(X)) -> a__half(mark(X)) , a__add(X1, X2) -> add(X1, X2) , a__add(s(X), Y) -> s(a__add(mark(X), mark(Y))) , a__add(0(), X) -> mark(X) , a__dbl(X) -> dbl(X) , a__dbl(s(X)) -> s(s(a__dbl(mark(X)))) , a__dbl(0()) -> 0() , 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() , a__half(X) -> half(X) , a__half(s(s(X))) -> s(a__half(mark(X))) , a__half(s(0())) -> 0() , a__half(0()) -> 0() , a__half(dbl(X)) -> mark(X) } 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^#(0()) -> c_5() , mark^#(0()) -> c_10() , mark^#(nil()) -> c_11() , a__add^#(X1, X2) -> c_17() , a__dbl^#(X) -> c_20() , a__dbl^#(0()) -> c_22() , a__first^#(X1, X2) -> c_23() , a__first^#(0(), X) -> c_25() , a__half^#(X) -> c_26() , a__half^#(s(0())) -> c_28() , a__half^#(0()) -> c_29() } 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__sqr^#(s(X)) -> c_4(a__add^#(a__sqr(mark(X)), a__dbl(mark(X))), a__sqr^#(mark(X)), mark^#(X), a__dbl^#(mark(X)), mark^#(X)) , 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^#(X)) , mark^#(first(X1, X2)) -> c_12(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X)), mark^#(X)) , mark^#(sqr(X)) -> c_14(a__sqr^#(mark(X)), mark^#(X)) , mark^#(add(X1, X2)) -> c_15(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(half(X)) -> c_16(a__half^#(mark(X)), mark^#(X)) , a__add^#(s(X), Y) -> c_18(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , a__add^#(0(), X) -> c_19(mark^#(X)) , a__dbl^#(s(X)) -> c_21(a__dbl^#(mark(X)), mark^#(X)) , a__first^#(s(X), cons(Y, Z)) -> c_24(mark^#(Y)) , a__half^#(s(s(X))) -> c_27(a__half^#(mark(X)), mark^#(X)) , a__half^#(dbl(X)) -> c_30(mark^#(X)) } 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(a__add(a__sqr(mark(X)), a__dbl(mark(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(mark(X)) , mark(0()) -> 0() , mark(nil()) -> nil() , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2)) , mark(dbl(X)) -> a__dbl(mark(X)) , mark(sqr(X)) -> a__sqr(mark(X)) , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2)) , mark(half(X)) -> a__half(mark(X)) , a__add(X1, X2) -> add(X1, X2) , a__add(s(X), Y) -> s(a__add(mark(X), mark(Y))) , a__add(0(), X) -> mark(X) , a__dbl(X) -> dbl(X) , a__dbl(s(X)) -> s(s(a__dbl(mark(X)))) , a__dbl(0()) -> 0() , 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() , a__half(X) -> half(X) , a__half(s(s(X))) -> s(a__half(mark(X))) , a__half(s(0())) -> 0() , a__half(0()) -> 0() , a__half(dbl(X)) -> mark(X) } 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: Following exception was raised: stack overflow 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..