MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { minus(x, 0()) -> x , 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))) , le(0(), y) -> true() , le(s(x), 0()) -> false() , le(s(x), s(y)) -> le(x, y) , inc(0()) -> s(0()) , inc(s(x)) -> s(inc(x)) , log(x) -> logIter(x, 0()) , logIter(x, y) -> if(le(s(0()), x), le(s(s(0())), x), quot(x, s(s(0()))), inc(y)) , if(true(), true(), x, y) -> logIter(x, y) , if(true(), false(), x, s(y)) -> y , if(false(), b, x, y) -> logZeroError() } 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^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(0(), s(y)) -> c_3() , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) , le^#(0(), y) -> c_5() , le^#(s(x), 0()) -> c_6() , le^#(s(x), s(y)) -> c_7(le^#(x, y)) , inc^#(0()) -> c_8() , inc^#(s(x)) -> c_9(inc^#(x)) , log^#(x) -> c_10(logIter^#(x, 0())) , logIter^#(x, y) -> c_11(if^#(le(s(0()), x), le(s(s(0())), x), quot(x, s(s(0()))), inc(y))) , if^#(true(), true(), x, y) -> c_12(logIter^#(x, y)) , if^#(true(), false(), x, s(y)) -> c_13(y) , if^#(false(), b, x, y) -> c_14() } 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^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(0(), s(y)) -> c_3() , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) , le^#(0(), y) -> c_5() , le^#(s(x), 0()) -> c_6() , le^#(s(x), s(y)) -> c_7(le^#(x, y)) , inc^#(0()) -> c_8() , inc^#(s(x)) -> c_9(inc^#(x)) , log^#(x) -> c_10(logIter^#(x, 0())) , logIter^#(x, y) -> c_11(if^#(le(s(0()), x), le(s(s(0())), x), quot(x, s(s(0()))), inc(y))) , if^#(true(), true(), x, y) -> c_12(logIter^#(x, y)) , if^#(true(), false(), x, s(y)) -> c_13(y) , if^#(false(), b, x, y) -> c_14() } Strict Trs: { minus(x, 0()) -> x , 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))) , le(0(), y) -> true() , le(s(x), 0()) -> false() , le(s(x), s(y)) -> le(x, y) , inc(0()) -> s(0()) , inc(s(x)) -> s(inc(x)) , log(x) -> logIter(x, 0()) , logIter(x, y) -> if(le(s(0()), x), le(s(s(0())), x), quot(x, s(s(0()))), inc(y)) , if(true(), true(), x, y) -> logIter(x, y) , if(true(), false(), x, s(y)) -> y , if(false(), b, x, y) -> logZeroError() } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {3,5,6,8,14} by applications of Pre({3,5,6,8,14}) = {1,4,7,9,11,13}. Here rules are labeled as follows: DPs: { 1: minus^#(x, 0()) -> c_1(x) , 2: minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , 3: quot^#(0(), s(y)) -> c_3() , 4: quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) , 5: le^#(0(), y) -> c_5() , 6: le^#(s(x), 0()) -> c_6() , 7: le^#(s(x), s(y)) -> c_7(le^#(x, y)) , 8: inc^#(0()) -> c_8() , 9: inc^#(s(x)) -> c_9(inc^#(x)) , 10: log^#(x) -> c_10(logIter^#(x, 0())) , 11: logIter^#(x, y) -> c_11(if^#(le(s(0()), x), le(s(s(0())), x), quot(x, s(s(0()))), inc(y))) , 12: if^#(true(), true(), x, y) -> c_12(logIter^#(x, y)) , 13: if^#(true(), false(), x, s(y)) -> c_13(y) , 14: if^#(false(), b, x, y) -> c_14() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { minus^#(x, 0()) -> c_1(x) , minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) , le^#(s(x), s(y)) -> c_7(le^#(x, y)) , inc^#(s(x)) -> c_9(inc^#(x)) , log^#(x) -> c_10(logIter^#(x, 0())) , logIter^#(x, y) -> c_11(if^#(le(s(0()), x), le(s(s(0())), x), quot(x, s(s(0()))), inc(y))) , if^#(true(), true(), x, y) -> c_12(logIter^#(x, y)) , if^#(true(), false(), x, s(y)) -> c_13(y) } Strict Trs: { minus(x, 0()) -> x , 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))) , le(0(), y) -> true() , le(s(x), 0()) -> false() , le(s(x), s(y)) -> le(x, y) , inc(0()) -> s(0()) , inc(s(x)) -> s(inc(x)) , log(x) -> logIter(x, 0()) , logIter(x, y) -> if(le(s(0()), x), le(s(s(0())), x), quot(x, s(s(0()))), inc(y)) , if(true(), true(), x, y) -> logIter(x, y) , if(true(), false(), x, s(y)) -> y , if(false(), b, x, y) -> logZeroError() } Weak DPs: { quot^#(0(), s(y)) -> c_3() , le^#(0(), y) -> c_5() , le^#(s(x), 0()) -> c_6() , inc^#(0()) -> c_8() , if^#(false(), b, x, y) -> c_14() } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. Arrrr..