MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { a__fact(X) -> a__if(a__zero(mark(X)), s(0()), prod(X, fact(p(X)))) , a__fact(X) -> fact(X) , a__if(X1, X2, X3) -> if(X1, X2, X3) , a__if(true(), X, Y) -> mark(X) , a__if(false(), X, Y) -> mark(Y) , a__zero(X) -> zero(X) , a__zero(s(X)) -> false() , a__zero(0()) -> true() , mark(s(X)) -> s(mark(X)) , mark(0()) -> 0() , mark(prod(X1, X2)) -> a__prod(mark(X1), mark(X2)) , mark(fact(X)) -> a__fact(mark(X)) , mark(p(X)) -> a__p(mark(X)) , mark(true()) -> true() , mark(false()) -> false() , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) , mark(zero(X)) -> a__zero(mark(X)) , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2)) , 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__prod(X1, X2) -> prod(X1, X2) , a__prod(s(X), Y) -> a__add(mark(Y), a__prod(mark(X), mark(Y))) , a__prod(0(), X) -> 0() , a__p(X) -> p(X) , a__p(s(X)) -> mark(X) } Obligation: innermost runtime complexity Answer: MAYBE We add following dependency tuples: Strict DPs: { a__fact^#(X) -> c_1(a__if^#(a__zero(mark(X)), s(0()), prod(X, fact(p(X)))), a__zero^#(mark(X)), mark^#(X)) , a__fact^#(X) -> c_2() , a__if^#(X1, X2, X3) -> c_3() , a__if^#(true(), X, Y) -> c_4(mark^#(X)) , a__if^#(false(), X, Y) -> c_5(mark^#(Y)) , a__zero^#(X) -> c_6() , a__zero^#(s(X)) -> c_7() , a__zero^#(0()) -> c_8() , mark^#(s(X)) -> c_9(mark^#(X)) , mark^#(0()) -> c_10() , mark^#(prod(X1, X2)) -> c_11(a__prod^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(fact(X)) -> c_12(a__fact^#(mark(X)), mark^#(X)) , mark^#(p(X)) -> c_13(a__p^#(mark(X)), mark^#(X)) , mark^#(true()) -> c_14() , mark^#(false()) -> c_15() , mark^#(if(X1, X2, X3)) -> c_16(a__if^#(mark(X1), X2, X3), mark^#(X1)) , mark^#(zero(X)) -> c_17(a__zero^#(mark(X)), mark^#(X)) , mark^#(add(X1, X2)) -> c_18(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , a__prod^#(X1, X2) -> c_22() , a__prod^#(s(X), Y) -> c_23(a__add^#(mark(Y), a__prod(mark(X), mark(Y))), mark^#(Y), a__prod^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , a__prod^#(0(), X) -> c_24() , a__p^#(X) -> c_25() , a__p^#(s(X)) -> c_26(mark^#(X)) , a__add^#(X1, X2) -> c_19() , a__add^#(s(X), Y) -> c_20(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , a__add^#(0(), X) -> c_21(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__fact^#(X) -> c_1(a__if^#(a__zero(mark(X)), s(0()), prod(X, fact(p(X)))), a__zero^#(mark(X)), mark^#(X)) , a__fact^#(X) -> c_2() , a__if^#(X1, X2, X3) -> c_3() , a__if^#(true(), X, Y) -> c_4(mark^#(X)) , a__if^#(false(), X, Y) -> c_5(mark^#(Y)) , a__zero^#(X) -> c_6() , a__zero^#(s(X)) -> c_7() , a__zero^#(0()) -> c_8() , mark^#(s(X)) -> c_9(mark^#(X)) , mark^#(0()) -> c_10() , mark^#(prod(X1, X2)) -> c_11(a__prod^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(fact(X)) -> c_12(a__fact^#(mark(X)), mark^#(X)) , mark^#(p(X)) -> c_13(a__p^#(mark(X)), mark^#(X)) , mark^#(true()) -> c_14() , mark^#(false()) -> c_15() , mark^#(if(X1, X2, X3)) -> c_16(a__if^#(mark(X1), X2, X3), mark^#(X1)) , mark^#(zero(X)) -> c_17(a__zero^#(mark(X)), mark^#(X)) , mark^#(add(X1, X2)) -> c_18(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , a__prod^#(X1, X2) -> c_22() , a__prod^#(s(X), Y) -> c_23(a__add^#(mark(Y), a__prod(mark(X), mark(Y))), mark^#(Y), a__prod^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , a__prod^#(0(), X) -> c_24() , a__p^#(X) -> c_25() , a__p^#(s(X)) -> c_26(mark^#(X)) , a__add^#(X1, X2) -> c_19() , a__add^#(s(X), Y) -> c_20(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , a__add^#(0(), X) -> c_21(mark^#(X)) } Weak Trs: { a__fact(X) -> a__if(a__zero(mark(X)), s(0()), prod(X, fact(p(X)))) , a__fact(X) -> fact(X) , a__if(X1, X2, X3) -> if(X1, X2, X3) , a__if(true(), X, Y) -> mark(X) , a__if(false(), X, Y) -> mark(Y) , a__zero(X) -> zero(X) , a__zero(s(X)) -> false() , a__zero(0()) -> true() , mark(s(X)) -> s(mark(X)) , mark(0()) -> 0() , mark(prod(X1, X2)) -> a__prod(mark(X1), mark(X2)) , mark(fact(X)) -> a__fact(mark(X)) , mark(p(X)) -> a__p(mark(X)) , mark(true()) -> true() , mark(false()) -> false() , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) , mark(zero(X)) -> a__zero(mark(X)) , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2)) , 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__prod(X1, X2) -> prod(X1, X2) , a__prod(s(X), Y) -> a__add(mark(Y), a__prod(mark(X), mark(Y))) , a__prod(0(), X) -> 0() , a__p(X) -> p(X) , a__p(s(X)) -> mark(X) } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {2,3,6,7,8,10,14,15,19,21,22,24} by applications of Pre({2,3,6,7,8,10,14,15,19,21,22,24}) = {1,4,5,9,11,12,13,16,17,18,20,23,25,26}. Here rules are labeled as follows: DPs: { 1: a__fact^#(X) -> c_1(a__if^#(a__zero(mark(X)), s(0()), prod(X, fact(p(X)))), a__zero^#(mark(X)), mark^#(X)) , 2: a__fact^#(X) -> c_2() , 3: a__if^#(X1, X2, X3) -> c_3() , 4: a__if^#(true(), X, Y) -> c_4(mark^#(X)) , 5: a__if^#(false(), X, Y) -> c_5(mark^#(Y)) , 6: a__zero^#(X) -> c_6() , 7: a__zero^#(s(X)) -> c_7() , 8: a__zero^#(0()) -> c_8() , 9: mark^#(s(X)) -> c_9(mark^#(X)) , 10: mark^#(0()) -> c_10() , 11: mark^#(prod(X1, X2)) -> c_11(a__prod^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , 12: mark^#(fact(X)) -> c_12(a__fact^#(mark(X)), mark^#(X)) , 13: mark^#(p(X)) -> c_13(a__p^#(mark(X)), mark^#(X)) , 14: mark^#(true()) -> c_14() , 15: mark^#(false()) -> c_15() , 16: mark^#(if(X1, X2, X3)) -> c_16(a__if^#(mark(X1), X2, X3), mark^#(X1)) , 17: mark^#(zero(X)) -> c_17(a__zero^#(mark(X)), mark^#(X)) , 18: mark^#(add(X1, X2)) -> c_18(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , 19: a__prod^#(X1, X2) -> c_22() , 20: a__prod^#(s(X), Y) -> c_23(a__add^#(mark(Y), a__prod(mark(X), mark(Y))), mark^#(Y), a__prod^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , 21: a__prod^#(0(), X) -> c_24() , 22: a__p^#(X) -> c_25() , 23: a__p^#(s(X)) -> c_26(mark^#(X)) , 24: a__add^#(X1, X2) -> c_19() , 25: a__add^#(s(X), Y) -> c_20(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , 26: a__add^#(0(), X) -> c_21(mark^#(X)) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__fact^#(X) -> c_1(a__if^#(a__zero(mark(X)), s(0()), prod(X, fact(p(X)))), a__zero^#(mark(X)), mark^#(X)) , a__if^#(true(), X, Y) -> c_4(mark^#(X)) , a__if^#(false(), X, Y) -> c_5(mark^#(Y)) , mark^#(s(X)) -> c_9(mark^#(X)) , mark^#(prod(X1, X2)) -> c_11(a__prod^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(fact(X)) -> c_12(a__fact^#(mark(X)), mark^#(X)) , mark^#(p(X)) -> c_13(a__p^#(mark(X)), mark^#(X)) , mark^#(if(X1, X2, X3)) -> c_16(a__if^#(mark(X1), X2, X3), mark^#(X1)) , mark^#(zero(X)) -> c_17(a__zero^#(mark(X)), mark^#(X)) , mark^#(add(X1, X2)) -> c_18(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , a__prod^#(s(X), Y) -> c_23(a__add^#(mark(Y), a__prod(mark(X), mark(Y))), mark^#(Y), a__prod^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , a__p^#(s(X)) -> c_26(mark^#(X)) , a__add^#(s(X), Y) -> c_20(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , a__add^#(0(), X) -> c_21(mark^#(X)) } Weak DPs: { a__fact^#(X) -> c_2() , a__if^#(X1, X2, X3) -> c_3() , a__zero^#(X) -> c_6() , a__zero^#(s(X)) -> c_7() , a__zero^#(0()) -> c_8() , mark^#(0()) -> c_10() , mark^#(true()) -> c_14() , mark^#(false()) -> c_15() , a__prod^#(X1, X2) -> c_22() , a__prod^#(0(), X) -> c_24() , a__p^#(X) -> c_25() , a__add^#(X1, X2) -> c_19() } Weak Trs: { a__fact(X) -> a__if(a__zero(mark(X)), s(0()), prod(X, fact(p(X)))) , a__fact(X) -> fact(X) , a__if(X1, X2, X3) -> if(X1, X2, X3) , a__if(true(), X, Y) -> mark(X) , a__if(false(), X, Y) -> mark(Y) , a__zero(X) -> zero(X) , a__zero(s(X)) -> false() , a__zero(0()) -> true() , mark(s(X)) -> s(mark(X)) , mark(0()) -> 0() , mark(prod(X1, X2)) -> a__prod(mark(X1), mark(X2)) , mark(fact(X)) -> a__fact(mark(X)) , mark(p(X)) -> a__p(mark(X)) , mark(true()) -> true() , mark(false()) -> false() , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) , mark(zero(X)) -> a__zero(mark(X)) , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2)) , 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__prod(X1, X2) -> prod(X1, X2) , a__prod(s(X), Y) -> a__add(mark(Y), a__prod(mark(X), mark(Y))) , a__prod(0(), X) -> 0() , a__p(X) -> p(X) , a__p(s(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__fact^#(X) -> c_2() , a__if^#(X1, X2, X3) -> c_3() , a__zero^#(X) -> c_6() , a__zero^#(s(X)) -> c_7() , a__zero^#(0()) -> c_8() , mark^#(0()) -> c_10() , mark^#(true()) -> c_14() , mark^#(false()) -> c_15() , a__prod^#(X1, X2) -> c_22() , a__prod^#(0(), X) -> c_24() , a__p^#(X) -> c_25() , a__add^#(X1, X2) -> c_19() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__fact^#(X) -> c_1(a__if^#(a__zero(mark(X)), s(0()), prod(X, fact(p(X)))), a__zero^#(mark(X)), mark^#(X)) , a__if^#(true(), X, Y) -> c_4(mark^#(X)) , a__if^#(false(), X, Y) -> c_5(mark^#(Y)) , mark^#(s(X)) -> c_9(mark^#(X)) , mark^#(prod(X1, X2)) -> c_11(a__prod^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(fact(X)) -> c_12(a__fact^#(mark(X)), mark^#(X)) , mark^#(p(X)) -> c_13(a__p^#(mark(X)), mark^#(X)) , mark^#(if(X1, X2, X3)) -> c_16(a__if^#(mark(X1), X2, X3), mark^#(X1)) , mark^#(zero(X)) -> c_17(a__zero^#(mark(X)), mark^#(X)) , mark^#(add(X1, X2)) -> c_18(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , a__prod^#(s(X), Y) -> c_23(a__add^#(mark(Y), a__prod(mark(X), mark(Y))), mark^#(Y), a__prod^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , a__p^#(s(X)) -> c_26(mark^#(X)) , a__add^#(s(X), Y) -> c_20(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , a__add^#(0(), X) -> c_21(mark^#(X)) } Weak Trs: { a__fact(X) -> a__if(a__zero(mark(X)), s(0()), prod(X, fact(p(X)))) , a__fact(X) -> fact(X) , a__if(X1, X2, X3) -> if(X1, X2, X3) , a__if(true(), X, Y) -> mark(X) , a__if(false(), X, Y) -> mark(Y) , a__zero(X) -> zero(X) , a__zero(s(X)) -> false() , a__zero(0()) -> true() , mark(s(X)) -> s(mark(X)) , mark(0()) -> 0() , mark(prod(X1, X2)) -> a__prod(mark(X1), mark(X2)) , mark(fact(X)) -> a__fact(mark(X)) , mark(p(X)) -> a__p(mark(X)) , mark(true()) -> true() , mark(false()) -> false() , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) , mark(zero(X)) -> a__zero(mark(X)) , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2)) , 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__prod(X1, X2) -> prod(X1, X2) , a__prod(s(X), Y) -> a__add(mark(Y), a__prod(mark(X), mark(Y))) , a__prod(0(), X) -> 0() , a__p(X) -> p(X) , a__p(s(X)) -> mark(X) } 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__fact^#(X) -> c_1(a__if^#(a__zero(mark(X)), s(0()), prod(X, fact(p(X)))), a__zero^#(mark(X)), mark^#(X)) , mark^#(zero(X)) -> c_17(a__zero^#(mark(X)), mark^#(X)) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__fact^#(X) -> c_1(a__if^#(a__zero(mark(X)), s(0()), prod(X, fact(p(X)))), mark^#(X)) , a__if^#(true(), X, Y) -> c_2(mark^#(X)) , a__if^#(false(), X, Y) -> c_3(mark^#(Y)) , mark^#(s(X)) -> c_4(mark^#(X)) , mark^#(prod(X1, X2)) -> c_5(a__prod^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(fact(X)) -> c_6(a__fact^#(mark(X)), mark^#(X)) , mark^#(p(X)) -> c_7(a__p^#(mark(X)), mark^#(X)) , mark^#(if(X1, X2, X3)) -> c_8(a__if^#(mark(X1), X2, X3), mark^#(X1)) , mark^#(zero(X)) -> c_9(mark^#(X)) , mark^#(add(X1, X2)) -> c_10(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , a__prod^#(s(X), Y) -> c_11(a__add^#(mark(Y), a__prod(mark(X), mark(Y))), mark^#(Y), a__prod^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , a__p^#(s(X)) -> c_12(mark^#(X)) , a__add^#(s(X), Y) -> c_13(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y)) , a__add^#(0(), X) -> c_14(mark^#(X)) } Weak Trs: { a__fact(X) -> a__if(a__zero(mark(X)), s(0()), prod(X, fact(p(X)))) , a__fact(X) -> fact(X) , a__if(X1, X2, X3) -> if(X1, X2, X3) , a__if(true(), X, Y) -> mark(X) , a__if(false(), X, Y) -> mark(Y) , a__zero(X) -> zero(X) , a__zero(s(X)) -> false() , a__zero(0()) -> true() , mark(s(X)) -> s(mark(X)) , mark(0()) -> 0() , mark(prod(X1, X2)) -> a__prod(mark(X1), mark(X2)) , mark(fact(X)) -> a__fact(mark(X)) , mark(p(X)) -> a__p(mark(X)) , mark(true()) -> true() , mark(false()) -> false() , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) , mark(zero(X)) -> a__zero(mark(X)) , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2)) , 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__prod(X1, X2) -> prod(X1, X2) , a__prod(s(X), Y) -> a__add(mark(Y), a__prod(mark(X), mark(Y))) , a__prod(0(), X) -> 0() , a__p(X) -> p(X) , a__p(s(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..