MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { le(0(), y) -> true() , le(s(x), 0()) -> false() , le(s(x), s(y)) -> le(x, y) , inc(0()) -> 0() , inc(s(x)) -> s(inc(x)) , minus(x, 0()) -> x , minus(0(), y) -> 0() , 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))) , log(x) -> log2(x, 0()) , log2(x, y) -> if(le(x, 0()), le(x, s(0())), x, inc(y)) , if(true(), b, x, y) -> log_undefined() , if(false(), b, x, y) -> if2(b, x, y) , if2(true(), x, s(y)) -> y , if2(false(), x, y) -> log2(quot(x, s(s(0()))), y) } 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) '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 minimal-enrichment and initial automaton 'match'' failed due to the following reason: match-boundness of the problem could not be verified. 2) 'Bounds with perSymbol-enrichment and initial automaton 'match'' failed due to the following reason: match-boundness of the problem could not be verified. 3) '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) 'Innermost Weak Dependency Pairs (timeout of 60 seconds)' failed due to the following reason: We add the following weak dependency pairs: Strict DPs: { le^#(0(), y) -> c_1() , le^#(s(x), 0()) -> c_2() , le^#(s(x), s(y)) -> c_3(le^#(x, y)) , inc^#(0()) -> c_4() , inc^#(s(x)) -> c_5(inc^#(x)) , minus^#(x, 0()) -> c_6(x) , minus^#(0(), y) -> c_7() , minus^#(s(x), s(y)) -> c_8(minus^#(x, y)) , quot^#(0(), s(y)) -> c_9() , quot^#(s(x), s(y)) -> c_10(quot^#(minus(x, y), s(y))) , log^#(x) -> c_11(log2^#(x, 0())) , log2^#(x, y) -> c_12(if^#(le(x, 0()), le(x, s(0())), x, inc(y))) , if^#(true(), b, x, y) -> c_13() , if^#(false(), b, x, y) -> c_14(if2^#(b, x, y)) , if2^#(true(), x, s(y)) -> c_15(y) , if2^#(false(), x, y) -> c_16(log2^#(quot(x, s(s(0()))), y)) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { le^#(0(), y) -> c_1() , le^#(s(x), 0()) -> c_2() , le^#(s(x), s(y)) -> c_3(le^#(x, y)) , inc^#(0()) -> c_4() , inc^#(s(x)) -> c_5(inc^#(x)) , minus^#(x, 0()) -> c_6(x) , minus^#(0(), y) -> c_7() , minus^#(s(x), s(y)) -> c_8(minus^#(x, y)) , quot^#(0(), s(y)) -> c_9() , quot^#(s(x), s(y)) -> c_10(quot^#(minus(x, y), s(y))) , log^#(x) -> c_11(log2^#(x, 0())) , log2^#(x, y) -> c_12(if^#(le(x, 0()), le(x, s(0())), x, inc(y))) , if^#(true(), b, x, y) -> c_13() , if^#(false(), b, x, y) -> c_14(if2^#(b, x, y)) , if2^#(true(), x, s(y)) -> c_15(y) , if2^#(false(), x, y) -> c_16(log2^#(quot(x, s(s(0()))), y)) } Strict Trs: { le(0(), y) -> true() , le(s(x), 0()) -> false() , le(s(x), s(y)) -> le(x, y) , inc(0()) -> 0() , inc(s(x)) -> s(inc(x)) , minus(x, 0()) -> x , minus(0(), y) -> 0() , 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))) , log(x) -> log2(x, 0()) , log2(x, y) -> if(le(x, 0()), le(x, s(0())), x, inc(y)) , if(true(), b, x, y) -> log_undefined() , if(false(), b, x, y) -> if2(b, x, y) , if2(true(), x, s(y)) -> y , if2(false(), x, y) -> log2(quot(x, s(s(0()))), y) } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {1,2,4,7,9,13} by applications of Pre({1,2,4,7,9,13}) = {3,5,6,8,10,12,15}. Here rules are labeled as follows: DPs: { 1: le^#(0(), y) -> c_1() , 2: le^#(s(x), 0()) -> c_2() , 3: le^#(s(x), s(y)) -> c_3(le^#(x, y)) , 4: inc^#(0()) -> c_4() , 5: inc^#(s(x)) -> c_5(inc^#(x)) , 6: minus^#(x, 0()) -> c_6(x) , 7: minus^#(0(), y) -> c_7() , 8: minus^#(s(x), s(y)) -> c_8(minus^#(x, y)) , 9: quot^#(0(), s(y)) -> c_9() , 10: quot^#(s(x), s(y)) -> c_10(quot^#(minus(x, y), s(y))) , 11: log^#(x) -> c_11(log2^#(x, 0())) , 12: log2^#(x, y) -> c_12(if^#(le(x, 0()), le(x, s(0())), x, inc(y))) , 13: if^#(true(), b, x, y) -> c_13() , 14: if^#(false(), b, x, y) -> c_14(if2^#(b, x, y)) , 15: if2^#(true(), x, s(y)) -> c_15(y) , 16: if2^#(false(), x, y) -> c_16(log2^#(quot(x, s(s(0()))), y)) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { le^#(s(x), s(y)) -> c_3(le^#(x, y)) , inc^#(s(x)) -> c_5(inc^#(x)) , minus^#(x, 0()) -> c_6(x) , minus^#(s(x), s(y)) -> c_8(minus^#(x, y)) , quot^#(s(x), s(y)) -> c_10(quot^#(minus(x, y), s(y))) , log^#(x) -> c_11(log2^#(x, 0())) , log2^#(x, y) -> c_12(if^#(le(x, 0()), le(x, s(0())), x, inc(y))) , if^#(false(), b, x, y) -> c_14(if2^#(b, x, y)) , if2^#(true(), x, s(y)) -> c_15(y) , if2^#(false(), x, y) -> c_16(log2^#(quot(x, s(s(0()))), y)) } Strict Trs: { le(0(), y) -> true() , le(s(x), 0()) -> false() , le(s(x), s(y)) -> le(x, y) , inc(0()) -> 0() , inc(s(x)) -> s(inc(x)) , minus(x, 0()) -> x , minus(0(), y) -> 0() , 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))) , log(x) -> log2(x, 0()) , log2(x, y) -> if(le(x, 0()), le(x, s(0())), x, inc(y)) , if(true(), b, x, y) -> log_undefined() , if(false(), b, x, y) -> if2(b, x, y) , if2(true(), x, s(y)) -> y , if2(false(), x, y) -> log2(quot(x, s(s(0()))), y) } Weak DPs: { le^#(0(), y) -> c_1() , le^#(s(x), 0()) -> c_2() , inc^#(0()) -> c_4() , minus^#(0(), y) -> c_7() , quot^#(0(), s(y)) -> c_9() , if^#(true(), b, x, y) -> c_13() } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. Arrrr..