MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { 0(#()) -> #() , +(x, #()) -> x , +(0(x), 0(y)) -> 0(+(x, y)) , +(0(x), 1(y)) -> 1(+(x, y)) , +(#(), x) -> x , +(+(x, y), z) -> +(x, +(y, z)) , +(1(x), 0(y)) -> 1(+(x, y)) , +(1(x), 1(y)) -> 0(+(+(x, y), 1(#()))) , *(x, +(y, z)) -> +(*(x, y), *(x, z)) , *(0(x), y) -> 0(*(x, y)) , *(#(), x) -> #() , *(1(x), y) -> +(0(*(x, y)), y) , *(*(x, y), z) -> *(x, *(y, z)) , app(nil(), l) -> l , app(cons(x, l1), l2) -> cons(x, app(l1, l2)) , sum(app(l1, l2)) -> +(sum(l1), sum(l2)) , sum(nil()) -> 0(#()) , sum(cons(x, l)) -> +(x, sum(l)) , prod(app(l1, l2)) -> *(prod(l1), prod(l2)) , prod(nil()) -> 1(#()) , prod(cons(x, l)) -> *(x, prod(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: { 0^#(#()) -> c_1() , +^#(x, #()) -> c_2(x) , +^#(0(x), 0(y)) -> c_3(0^#(+(x, y))) , +^#(0(x), 1(y)) -> c_4(+^#(x, y)) , +^#(#(), x) -> c_5(x) , +^#(+(x, y), z) -> c_6(+^#(x, +(y, z))) , +^#(1(x), 0(y)) -> c_7(+^#(x, y)) , +^#(1(x), 1(y)) -> c_8(0^#(+(+(x, y), 1(#())))) , *^#(x, +(y, z)) -> c_9(+^#(*(x, y), *(x, z))) , *^#(0(x), y) -> c_10(0^#(*(x, y))) , *^#(#(), x) -> c_11() , *^#(1(x), y) -> c_12(+^#(0(*(x, y)), y)) , *^#(*(x, y), z) -> c_13(*^#(x, *(y, z))) , app^#(nil(), l) -> c_14(l) , app^#(cons(x, l1), l2) -> c_15(x, app^#(l1, l2)) , sum^#(app(l1, l2)) -> c_16(+^#(sum(l1), sum(l2))) , sum^#(nil()) -> c_17(0^#(#())) , sum^#(cons(x, l)) -> c_18(+^#(x, sum(l))) , prod^#(app(l1, l2)) -> c_19(*^#(prod(l1), prod(l2))) , prod^#(nil()) -> c_20() , prod^#(cons(x, l)) -> c_21(*^#(x, prod(l))) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { 0^#(#()) -> c_1() , +^#(x, #()) -> c_2(x) , +^#(0(x), 0(y)) -> c_3(0^#(+(x, y))) , +^#(0(x), 1(y)) -> c_4(+^#(x, y)) , +^#(#(), x) -> c_5(x) , +^#(+(x, y), z) -> c_6(+^#(x, +(y, z))) , +^#(1(x), 0(y)) -> c_7(+^#(x, y)) , +^#(1(x), 1(y)) -> c_8(0^#(+(+(x, y), 1(#())))) , *^#(x, +(y, z)) -> c_9(+^#(*(x, y), *(x, z))) , *^#(0(x), y) -> c_10(0^#(*(x, y))) , *^#(#(), x) -> c_11() , *^#(1(x), y) -> c_12(+^#(0(*(x, y)), y)) , *^#(*(x, y), z) -> c_13(*^#(x, *(y, z))) , app^#(nil(), l) -> c_14(l) , app^#(cons(x, l1), l2) -> c_15(x, app^#(l1, l2)) , sum^#(app(l1, l2)) -> c_16(+^#(sum(l1), sum(l2))) , sum^#(nil()) -> c_17(0^#(#())) , sum^#(cons(x, l)) -> c_18(+^#(x, sum(l))) , prod^#(app(l1, l2)) -> c_19(*^#(prod(l1), prod(l2))) , prod^#(nil()) -> c_20() , prod^#(cons(x, l)) -> c_21(*^#(x, prod(l))) } Strict Trs: { 0(#()) -> #() , +(x, #()) -> x , +(0(x), 0(y)) -> 0(+(x, y)) , +(0(x), 1(y)) -> 1(+(x, y)) , +(#(), x) -> x , +(+(x, y), z) -> +(x, +(y, z)) , +(1(x), 0(y)) -> 1(+(x, y)) , +(1(x), 1(y)) -> 0(+(+(x, y), 1(#()))) , *(x, +(y, z)) -> +(*(x, y), *(x, z)) , *(0(x), y) -> 0(*(x, y)) , *(#(), x) -> #() , *(1(x), y) -> +(0(*(x, y)), y) , *(*(x, y), z) -> *(x, *(y, z)) , app(nil(), l) -> l , app(cons(x, l1), l2) -> cons(x, app(l1, l2)) , sum(app(l1, l2)) -> +(sum(l1), sum(l2)) , sum(nil()) -> 0(#()) , sum(cons(x, l)) -> +(x, sum(l)) , prod(app(l1, l2)) -> *(prod(l1), prod(l2)) , prod(nil()) -> 1(#()) , prod(cons(x, l)) -> *(x, prod(l)) } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {1,11,20} by applications of Pre({1,11,20}) = {2,3,5,8,10,13,14,15,17,19,21}. Here rules are labeled as follows: DPs: { 1: 0^#(#()) -> c_1() , 2: +^#(x, #()) -> c_2(x) , 3: +^#(0(x), 0(y)) -> c_3(0^#(+(x, y))) , 4: +^#(0(x), 1(y)) -> c_4(+^#(x, y)) , 5: +^#(#(), x) -> c_5(x) , 6: +^#(+(x, y), z) -> c_6(+^#(x, +(y, z))) , 7: +^#(1(x), 0(y)) -> c_7(+^#(x, y)) , 8: +^#(1(x), 1(y)) -> c_8(0^#(+(+(x, y), 1(#())))) , 9: *^#(x, +(y, z)) -> c_9(+^#(*(x, y), *(x, z))) , 10: *^#(0(x), y) -> c_10(0^#(*(x, y))) , 11: *^#(#(), x) -> c_11() , 12: *^#(1(x), y) -> c_12(+^#(0(*(x, y)), y)) , 13: *^#(*(x, y), z) -> c_13(*^#(x, *(y, z))) , 14: app^#(nil(), l) -> c_14(l) , 15: app^#(cons(x, l1), l2) -> c_15(x, app^#(l1, l2)) , 16: sum^#(app(l1, l2)) -> c_16(+^#(sum(l1), sum(l2))) , 17: sum^#(nil()) -> c_17(0^#(#())) , 18: sum^#(cons(x, l)) -> c_18(+^#(x, sum(l))) , 19: prod^#(app(l1, l2)) -> c_19(*^#(prod(l1), prod(l2))) , 20: prod^#(nil()) -> c_20() , 21: prod^#(cons(x, l)) -> c_21(*^#(x, prod(l))) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { +^#(x, #()) -> c_2(x) , +^#(0(x), 0(y)) -> c_3(0^#(+(x, y))) , +^#(0(x), 1(y)) -> c_4(+^#(x, y)) , +^#(#(), x) -> c_5(x) , +^#(+(x, y), z) -> c_6(+^#(x, +(y, z))) , +^#(1(x), 0(y)) -> c_7(+^#(x, y)) , +^#(1(x), 1(y)) -> c_8(0^#(+(+(x, y), 1(#())))) , *^#(x, +(y, z)) -> c_9(+^#(*(x, y), *(x, z))) , *^#(0(x), y) -> c_10(0^#(*(x, y))) , *^#(1(x), y) -> c_12(+^#(0(*(x, y)), y)) , *^#(*(x, y), z) -> c_13(*^#(x, *(y, z))) , app^#(nil(), l) -> c_14(l) , app^#(cons(x, l1), l2) -> c_15(x, app^#(l1, l2)) , sum^#(app(l1, l2)) -> c_16(+^#(sum(l1), sum(l2))) , sum^#(nil()) -> c_17(0^#(#())) , sum^#(cons(x, l)) -> c_18(+^#(x, sum(l))) , prod^#(app(l1, l2)) -> c_19(*^#(prod(l1), prod(l2))) , prod^#(cons(x, l)) -> c_21(*^#(x, prod(l))) } Strict Trs: { 0(#()) -> #() , +(x, #()) -> x , +(0(x), 0(y)) -> 0(+(x, y)) , +(0(x), 1(y)) -> 1(+(x, y)) , +(#(), x) -> x , +(+(x, y), z) -> +(x, +(y, z)) , +(1(x), 0(y)) -> 1(+(x, y)) , +(1(x), 1(y)) -> 0(+(+(x, y), 1(#()))) , *(x, +(y, z)) -> +(*(x, y), *(x, z)) , *(0(x), y) -> 0(*(x, y)) , *(#(), x) -> #() , *(1(x), y) -> +(0(*(x, y)), y) , *(*(x, y), z) -> *(x, *(y, z)) , app(nil(), l) -> l , app(cons(x, l1), l2) -> cons(x, app(l1, l2)) , sum(app(l1, l2)) -> +(sum(l1), sum(l2)) , sum(nil()) -> 0(#()) , sum(cons(x, l)) -> +(x, sum(l)) , prod(app(l1, l2)) -> *(prod(l1), prod(l2)) , prod(nil()) -> 1(#()) , prod(cons(x, l)) -> *(x, prod(l)) } Weak DPs: { 0^#(#()) -> c_1() , *^#(#(), x) -> c_11() , prod^#(nil()) -> c_20() } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {2,7,9,15} by applications of Pre({2,7,9,15}) = {1,3,4,5,6,8,10,11,12,13,14,16,17,18}. Here rules are labeled as follows: DPs: { 1: +^#(x, #()) -> c_2(x) , 2: +^#(0(x), 0(y)) -> c_3(0^#(+(x, y))) , 3: +^#(0(x), 1(y)) -> c_4(+^#(x, y)) , 4: +^#(#(), x) -> c_5(x) , 5: +^#(+(x, y), z) -> c_6(+^#(x, +(y, z))) , 6: +^#(1(x), 0(y)) -> c_7(+^#(x, y)) , 7: +^#(1(x), 1(y)) -> c_8(0^#(+(+(x, y), 1(#())))) , 8: *^#(x, +(y, z)) -> c_9(+^#(*(x, y), *(x, z))) , 9: *^#(0(x), y) -> c_10(0^#(*(x, y))) , 10: *^#(1(x), y) -> c_12(+^#(0(*(x, y)), y)) , 11: *^#(*(x, y), z) -> c_13(*^#(x, *(y, z))) , 12: app^#(nil(), l) -> c_14(l) , 13: app^#(cons(x, l1), l2) -> c_15(x, app^#(l1, l2)) , 14: sum^#(app(l1, l2)) -> c_16(+^#(sum(l1), sum(l2))) , 15: sum^#(nil()) -> c_17(0^#(#())) , 16: sum^#(cons(x, l)) -> c_18(+^#(x, sum(l))) , 17: prod^#(app(l1, l2)) -> c_19(*^#(prod(l1), prod(l2))) , 18: prod^#(cons(x, l)) -> c_21(*^#(x, prod(l))) , 19: 0^#(#()) -> c_1() , 20: *^#(#(), x) -> c_11() , 21: prod^#(nil()) -> c_20() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { +^#(x, #()) -> c_2(x) , +^#(0(x), 1(y)) -> c_4(+^#(x, y)) , +^#(#(), x) -> c_5(x) , +^#(+(x, y), z) -> c_6(+^#(x, +(y, z))) , +^#(1(x), 0(y)) -> c_7(+^#(x, y)) , *^#(x, +(y, z)) -> c_9(+^#(*(x, y), *(x, z))) , *^#(1(x), y) -> c_12(+^#(0(*(x, y)), y)) , *^#(*(x, y), z) -> c_13(*^#(x, *(y, z))) , app^#(nil(), l) -> c_14(l) , app^#(cons(x, l1), l2) -> c_15(x, app^#(l1, l2)) , sum^#(app(l1, l2)) -> c_16(+^#(sum(l1), sum(l2))) , sum^#(cons(x, l)) -> c_18(+^#(x, sum(l))) , prod^#(app(l1, l2)) -> c_19(*^#(prod(l1), prod(l2))) , prod^#(cons(x, l)) -> c_21(*^#(x, prod(l))) } Strict Trs: { 0(#()) -> #() , +(x, #()) -> x , +(0(x), 0(y)) -> 0(+(x, y)) , +(0(x), 1(y)) -> 1(+(x, y)) , +(#(), x) -> x , +(+(x, y), z) -> +(x, +(y, z)) , +(1(x), 0(y)) -> 1(+(x, y)) , +(1(x), 1(y)) -> 0(+(+(x, y), 1(#()))) , *(x, +(y, z)) -> +(*(x, y), *(x, z)) , *(0(x), y) -> 0(*(x, y)) , *(#(), x) -> #() , *(1(x), y) -> +(0(*(x, y)), y) , *(*(x, y), z) -> *(x, *(y, z)) , app(nil(), l) -> l , app(cons(x, l1), l2) -> cons(x, app(l1, l2)) , sum(app(l1, l2)) -> +(sum(l1), sum(l2)) , sum(nil()) -> 0(#()) , sum(cons(x, l)) -> +(x, sum(l)) , prod(app(l1, l2)) -> *(prod(l1), prod(l2)) , prod(nil()) -> 1(#()) , prod(cons(x, l)) -> *(x, prod(l)) } Weak DPs: { 0^#(#()) -> c_1() , +^#(0(x), 0(y)) -> c_3(0^#(+(x, y))) , +^#(1(x), 1(y)) -> c_8(0^#(+(+(x, y), 1(#())))) , *^#(0(x), y) -> c_10(0^#(*(x, y))) , *^#(#(), x) -> c_11() , sum^#(nil()) -> c_17(0^#(#())) , prod^#(nil()) -> c_20() } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. Arrrr..