MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { minus(x, 0()) -> x , minus(minus(x, y), z) -> minus(x, plus(y, z)) , minus(s(x), s(y)) -> minus(x, y) , quot(0(), s(y)) -> 0() , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , plus(0(), y) -> y , plus(s(x), y) -> s(plus(x, y)) , app(l, nil()) -> l , app(nil(), k) -> k , app(cons(x, l), k) -> cons(x, app(l, k)) , sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k))))) , sum(cons(x, nil())) -> cons(x, nil()) , sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l)) } 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) '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) '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) 'Innermost Weak Dependency Pairs (timeout of 60 seconds)' failed due to the following reason: We add the following weak dependency pairs: Strict DPs: { minus^#(x, 0()) -> c_1(x) , minus^#(minus(x, y), z) -> c_2(minus^#(x, plus(y, z))) , minus^#(s(x), s(y)) -> c_3(minus^#(x, y)) , quot^#(0(), s(y)) -> c_4() , quot^#(s(x), s(y)) -> c_5(quot^#(minus(x, y), s(y))) , plus^#(0(), y) -> c_6(y) , plus^#(s(x), y) -> c_7(plus^#(x, y)) , app^#(l, nil()) -> c_8(l) , app^#(nil(), k) -> c_9(k) , app^#(cons(x, l), k) -> c_10(x, app^#(l, k)) , sum^#(app(l, cons(x, cons(y, k)))) -> c_11(sum^#(app(l, sum(cons(x, cons(y, k)))))) , sum^#(cons(x, nil())) -> c_12(x) , sum^#(cons(x, cons(y, l))) -> c_13(sum^#(cons(plus(x, y), l))) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { minus^#(x, 0()) -> c_1(x) , minus^#(minus(x, y), z) -> c_2(minus^#(x, plus(y, z))) , minus^#(s(x), s(y)) -> c_3(minus^#(x, y)) , quot^#(0(), s(y)) -> c_4() , quot^#(s(x), s(y)) -> c_5(quot^#(minus(x, y), s(y))) , plus^#(0(), y) -> c_6(y) , plus^#(s(x), y) -> c_7(plus^#(x, y)) , app^#(l, nil()) -> c_8(l) , app^#(nil(), k) -> c_9(k) , app^#(cons(x, l), k) -> c_10(x, app^#(l, k)) , sum^#(app(l, cons(x, cons(y, k)))) -> c_11(sum^#(app(l, sum(cons(x, cons(y, k)))))) , sum^#(cons(x, nil())) -> c_12(x) , sum^#(cons(x, cons(y, l))) -> c_13(sum^#(cons(plus(x, y), l))) } Strict Trs: { minus(x, 0()) -> x , minus(minus(x, y), z) -> minus(x, plus(y, z)) , minus(s(x), s(y)) -> minus(x, y) , quot(0(), s(y)) -> 0() , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , plus(0(), y) -> y , plus(s(x), y) -> s(plus(x, y)) , app(l, nil()) -> l , app(nil(), k) -> k , app(cons(x, l), k) -> cons(x, app(l, k)) , sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k))))) , sum(cons(x, nil())) -> cons(x, nil()) , sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l)) } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {4} by applications of Pre({4}) = {1,5,6,8,9,10,12}. Here rules are labeled as follows: DPs: { 1: minus^#(x, 0()) -> c_1(x) , 2: minus^#(minus(x, y), z) -> c_2(minus^#(x, plus(y, z))) , 3: minus^#(s(x), s(y)) -> c_3(minus^#(x, y)) , 4: quot^#(0(), s(y)) -> c_4() , 5: quot^#(s(x), s(y)) -> c_5(quot^#(minus(x, y), s(y))) , 6: plus^#(0(), y) -> c_6(y) , 7: plus^#(s(x), y) -> c_7(plus^#(x, y)) , 8: app^#(l, nil()) -> c_8(l) , 9: app^#(nil(), k) -> c_9(k) , 10: app^#(cons(x, l), k) -> c_10(x, app^#(l, k)) , 11: sum^#(app(l, cons(x, cons(y, k)))) -> c_11(sum^#(app(l, sum(cons(x, cons(y, k)))))) , 12: sum^#(cons(x, nil())) -> c_12(x) , 13: sum^#(cons(x, cons(y, l))) -> c_13(sum^#(cons(plus(x, y), l))) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { minus^#(x, 0()) -> c_1(x) , minus^#(minus(x, y), z) -> c_2(minus^#(x, plus(y, z))) , minus^#(s(x), s(y)) -> c_3(minus^#(x, y)) , quot^#(s(x), s(y)) -> c_5(quot^#(minus(x, y), s(y))) , plus^#(0(), y) -> c_6(y) , plus^#(s(x), y) -> c_7(plus^#(x, y)) , app^#(l, nil()) -> c_8(l) , app^#(nil(), k) -> c_9(k) , app^#(cons(x, l), k) -> c_10(x, app^#(l, k)) , sum^#(app(l, cons(x, cons(y, k)))) -> c_11(sum^#(app(l, sum(cons(x, cons(y, k)))))) , sum^#(cons(x, nil())) -> c_12(x) , sum^#(cons(x, cons(y, l))) -> c_13(sum^#(cons(plus(x, y), l))) } Strict Trs: { minus(x, 0()) -> x , minus(minus(x, y), z) -> minus(x, plus(y, z)) , minus(s(x), s(y)) -> minus(x, y) , quot(0(), s(y)) -> 0() , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , plus(0(), y) -> y , plus(s(x), y) -> s(plus(x, y)) , app(l, nil()) -> l , app(nil(), k) -> k , app(cons(x, l), k) -> cons(x, app(l, k)) , sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k))))) , sum(cons(x, nil())) -> cons(x, nil()) , sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l)) } Weak DPs: { quot^#(0(), s(y)) -> c_4() } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. Arrrr..