MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { log(x, s(s(y))) -> cond(le(x, s(s(y))), x, y) , cond(true(), x, y) -> s(0()) , cond(false(), x, y) -> double(log(x, square(s(s(y))))) , le(s(u), s(v)) -> le(u, v) , le(s(u), 0()) -> false() , le(0(), v) -> true() , double(s(x)) -> s(s(double(x))) , double(0()) -> 0() , square(s(x)) -> s(plus(square(x), double(x))) , square(0()) -> 0() , plus(n, s(m)) -> s(plus(n, m)) , plus(n, 0()) -> n } Obligation: innermost runtime complexity Answer: MAYBE We add following dependency tuples: Strict DPs: { log^#(x, s(s(y))) -> c_1(cond^#(le(x, s(s(y))), x, y), le^#(x, s(s(y)))) , cond^#(true(), x, y) -> c_2() , cond^#(false(), x, y) -> c_3(double^#(log(x, square(s(s(y))))), log^#(x, square(s(s(y)))), square^#(s(s(y)))) , le^#(s(u), s(v)) -> c_4(le^#(u, v)) , le^#(s(u), 0()) -> c_5() , le^#(0(), v) -> c_6() , double^#(s(x)) -> c_7(double^#(x)) , double^#(0()) -> c_8() , square^#(s(x)) -> c_9(plus^#(square(x), double(x)), square^#(x), double^#(x)) , square^#(0()) -> c_10() , plus^#(n, s(m)) -> c_11(plus^#(n, m)) , plus^#(n, 0()) -> c_12() } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { log^#(x, s(s(y))) -> c_1(cond^#(le(x, s(s(y))), x, y), le^#(x, s(s(y)))) , cond^#(true(), x, y) -> c_2() , cond^#(false(), x, y) -> c_3(double^#(log(x, square(s(s(y))))), log^#(x, square(s(s(y)))), square^#(s(s(y)))) , le^#(s(u), s(v)) -> c_4(le^#(u, v)) , le^#(s(u), 0()) -> c_5() , le^#(0(), v) -> c_6() , double^#(s(x)) -> c_7(double^#(x)) , double^#(0()) -> c_8() , square^#(s(x)) -> c_9(plus^#(square(x), double(x)), square^#(x), double^#(x)) , square^#(0()) -> c_10() , plus^#(n, s(m)) -> c_11(plus^#(n, m)) , plus^#(n, 0()) -> c_12() } Weak Trs: { log(x, s(s(y))) -> cond(le(x, s(s(y))), x, y) , cond(true(), x, y) -> s(0()) , cond(false(), x, y) -> double(log(x, square(s(s(y))))) , le(s(u), s(v)) -> le(u, v) , le(s(u), 0()) -> false() , le(0(), v) -> true() , double(s(x)) -> s(s(double(x))) , double(0()) -> 0() , square(s(x)) -> s(plus(square(x), double(x))) , square(0()) -> 0() , plus(n, s(m)) -> s(plus(n, m)) , plus(n, 0()) -> n } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {2,5,6,8,10,12} by applications of Pre({2,5,6,8,10,12}) = {1,3,4,7,9,11}. Here rules are labeled as follows: DPs: { 1: log^#(x, s(s(y))) -> c_1(cond^#(le(x, s(s(y))), x, y), le^#(x, s(s(y)))) , 2: cond^#(true(), x, y) -> c_2() , 3: cond^#(false(), x, y) -> c_3(double^#(log(x, square(s(s(y))))), log^#(x, square(s(s(y)))), square^#(s(s(y)))) , 4: le^#(s(u), s(v)) -> c_4(le^#(u, v)) , 5: le^#(s(u), 0()) -> c_5() , 6: le^#(0(), v) -> c_6() , 7: double^#(s(x)) -> c_7(double^#(x)) , 8: double^#(0()) -> c_8() , 9: square^#(s(x)) -> c_9(plus^#(square(x), double(x)), square^#(x), double^#(x)) , 10: square^#(0()) -> c_10() , 11: plus^#(n, s(m)) -> c_11(plus^#(n, m)) , 12: plus^#(n, 0()) -> c_12() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { log^#(x, s(s(y))) -> c_1(cond^#(le(x, s(s(y))), x, y), le^#(x, s(s(y)))) , cond^#(false(), x, y) -> c_3(double^#(log(x, square(s(s(y))))), log^#(x, square(s(s(y)))), square^#(s(s(y)))) , le^#(s(u), s(v)) -> c_4(le^#(u, v)) , double^#(s(x)) -> c_7(double^#(x)) , square^#(s(x)) -> c_9(plus^#(square(x), double(x)), square^#(x), double^#(x)) , plus^#(n, s(m)) -> c_11(plus^#(n, m)) } Weak DPs: { cond^#(true(), x, y) -> c_2() , le^#(s(u), 0()) -> c_5() , le^#(0(), v) -> c_6() , double^#(0()) -> c_8() , square^#(0()) -> c_10() , plus^#(n, 0()) -> c_12() } Weak Trs: { log(x, s(s(y))) -> cond(le(x, s(s(y))), x, y) , cond(true(), x, y) -> s(0()) , cond(false(), x, y) -> double(log(x, square(s(s(y))))) , le(s(u), s(v)) -> le(u, v) , le(s(u), 0()) -> false() , le(0(), v) -> true() , double(s(x)) -> s(s(double(x))) , double(0()) -> 0() , square(s(x)) -> s(plus(square(x), double(x))) , square(0()) -> 0() , plus(n, s(m)) -> s(plus(n, m)) , plus(n, 0()) -> n } 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. { cond^#(true(), x, y) -> c_2() , le^#(s(u), 0()) -> c_5() , le^#(0(), v) -> c_6() , double^#(0()) -> c_8() , square^#(0()) -> c_10() , plus^#(n, 0()) -> c_12() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { log^#(x, s(s(y))) -> c_1(cond^#(le(x, s(s(y))), x, y), le^#(x, s(s(y)))) , cond^#(false(), x, y) -> c_3(double^#(log(x, square(s(s(y))))), log^#(x, square(s(s(y)))), square^#(s(s(y)))) , le^#(s(u), s(v)) -> c_4(le^#(u, v)) , double^#(s(x)) -> c_7(double^#(x)) , square^#(s(x)) -> c_9(plus^#(square(x), double(x)), square^#(x), double^#(x)) , plus^#(n, s(m)) -> c_11(plus^#(n, m)) } Weak Trs: { log(x, s(s(y))) -> cond(le(x, s(s(y))), x, y) , cond(true(), x, y) -> s(0()) , cond(false(), x, y) -> double(log(x, square(s(s(y))))) , le(s(u), s(v)) -> le(u, v) , le(s(u), 0()) -> false() , le(0(), v) -> true() , double(s(x)) -> s(s(double(x))) , double(0()) -> 0() , square(s(x)) -> s(plus(square(x), double(x))) , square(0()) -> 0() , plus(n, s(m)) -> s(plus(n, m)) , plus(n, 0()) -> n } Obligation: innermost runtime complexity Answer: MAYBE The input cannot be shown compatible Arrrr..