MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { if(true(), t, e) -> t , if(false(), t, e) -> e , member(x, nil()) -> false() , member(x, cons(y, ys)) -> if(eq(x, y), true(), member(x, ys)) , eq(nil(), nil()) -> true() , eq(O(x), 0(y)) -> eq(x, y) , eq(0(x), 1(y)) -> false() , eq(1(x), 0(y)) -> false() , eq(1(x), 1(y)) -> eq(x, y) , negate(0(x)) -> 1(x) , negate(1(x)) -> 0(x) , choice(cons(x, xs)) -> x , choice(cons(x, xs)) -> choice(xs) , guess(nil()) -> nil() , guess(cons(clause, cnf)) -> cons(choice(clause), guess(cnf)) , verify(nil()) -> true() , verify(cons(l, ls)) -> if(member(negate(l), ls), false(), verify(ls)) , sat(cnf) -> satck(cnf, guess(cnf)) , satck(cnf, assign) -> if(verify(assign), assign, unsat()) } 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: { if^#(true(), t, e) -> c_1(t) , if^#(false(), t, e) -> c_2(e) , member^#(x, nil()) -> c_3() , member^#(x, cons(y, ys)) -> c_4(if^#(eq(x, y), true(), member(x, ys))) , eq^#(nil(), nil()) -> c_5() , eq^#(O(x), 0(y)) -> c_6(eq^#(x, y)) , eq^#(0(x), 1(y)) -> c_7() , eq^#(1(x), 0(y)) -> c_8() , eq^#(1(x), 1(y)) -> c_9(eq^#(x, y)) , negate^#(0(x)) -> c_10(x) , negate^#(1(x)) -> c_11(x) , choice^#(cons(x, xs)) -> c_12(x) , choice^#(cons(x, xs)) -> c_13(choice^#(xs)) , guess^#(nil()) -> c_14() , guess^#(cons(clause, cnf)) -> c_15(choice^#(clause), guess^#(cnf)) , verify^#(nil()) -> c_16() , verify^#(cons(l, ls)) -> c_17(if^#(member(negate(l), ls), false(), verify(ls))) , sat^#(cnf) -> c_18(satck^#(cnf, guess(cnf))) , satck^#(cnf, assign) -> c_19(if^#(verify(assign), assign, unsat())) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { if^#(true(), t, e) -> c_1(t) , if^#(false(), t, e) -> c_2(e) , member^#(x, nil()) -> c_3() , member^#(x, cons(y, ys)) -> c_4(if^#(eq(x, y), true(), member(x, ys))) , eq^#(nil(), nil()) -> c_5() , eq^#(O(x), 0(y)) -> c_6(eq^#(x, y)) , eq^#(0(x), 1(y)) -> c_7() , eq^#(1(x), 0(y)) -> c_8() , eq^#(1(x), 1(y)) -> c_9(eq^#(x, y)) , negate^#(0(x)) -> c_10(x) , negate^#(1(x)) -> c_11(x) , choice^#(cons(x, xs)) -> c_12(x) , choice^#(cons(x, xs)) -> c_13(choice^#(xs)) , guess^#(nil()) -> c_14() , guess^#(cons(clause, cnf)) -> c_15(choice^#(clause), guess^#(cnf)) , verify^#(nil()) -> c_16() , verify^#(cons(l, ls)) -> c_17(if^#(member(negate(l), ls), false(), verify(ls))) , sat^#(cnf) -> c_18(satck^#(cnf, guess(cnf))) , satck^#(cnf, assign) -> c_19(if^#(verify(assign), assign, unsat())) } Strict Trs: { if(true(), t, e) -> t , if(false(), t, e) -> e , member(x, nil()) -> false() , member(x, cons(y, ys)) -> if(eq(x, y), true(), member(x, ys)) , eq(nil(), nil()) -> true() , eq(O(x), 0(y)) -> eq(x, y) , eq(0(x), 1(y)) -> false() , eq(1(x), 0(y)) -> false() , eq(1(x), 1(y)) -> eq(x, y) , negate(0(x)) -> 1(x) , negate(1(x)) -> 0(x) , choice(cons(x, xs)) -> x , choice(cons(x, xs)) -> choice(xs) , guess(nil()) -> nil() , guess(cons(clause, cnf)) -> cons(choice(clause), guess(cnf)) , verify(nil()) -> true() , verify(cons(l, ls)) -> if(member(negate(l), ls), false(), verify(ls)) , sat(cnf) -> satck(cnf, guess(cnf)) , satck(cnf, assign) -> if(verify(assign), assign, unsat()) } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {3,5,7,8,14,16} by applications of Pre({3,5,7,8,14,16}) = {1,2,6,9,10,11,12,15}. Here rules are labeled as follows: DPs: { 1: if^#(true(), t, e) -> c_1(t) , 2: if^#(false(), t, e) -> c_2(e) , 3: member^#(x, nil()) -> c_3() , 4: member^#(x, cons(y, ys)) -> c_4(if^#(eq(x, y), true(), member(x, ys))) , 5: eq^#(nil(), nil()) -> c_5() , 6: eq^#(O(x), 0(y)) -> c_6(eq^#(x, y)) , 7: eq^#(0(x), 1(y)) -> c_7() , 8: eq^#(1(x), 0(y)) -> c_8() , 9: eq^#(1(x), 1(y)) -> c_9(eq^#(x, y)) , 10: negate^#(0(x)) -> c_10(x) , 11: negate^#(1(x)) -> c_11(x) , 12: choice^#(cons(x, xs)) -> c_12(x) , 13: choice^#(cons(x, xs)) -> c_13(choice^#(xs)) , 14: guess^#(nil()) -> c_14() , 15: guess^#(cons(clause, cnf)) -> c_15(choice^#(clause), guess^#(cnf)) , 16: verify^#(nil()) -> c_16() , 17: verify^#(cons(l, ls)) -> c_17(if^#(member(negate(l), ls), false(), verify(ls))) , 18: sat^#(cnf) -> c_18(satck^#(cnf, guess(cnf))) , 19: satck^#(cnf, assign) -> c_19(if^#(verify(assign), assign, unsat())) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { if^#(true(), t, e) -> c_1(t) , if^#(false(), t, e) -> c_2(e) , member^#(x, cons(y, ys)) -> c_4(if^#(eq(x, y), true(), member(x, ys))) , eq^#(O(x), 0(y)) -> c_6(eq^#(x, y)) , eq^#(1(x), 1(y)) -> c_9(eq^#(x, y)) , negate^#(0(x)) -> c_10(x) , negate^#(1(x)) -> c_11(x) , choice^#(cons(x, xs)) -> c_12(x) , choice^#(cons(x, xs)) -> c_13(choice^#(xs)) , guess^#(cons(clause, cnf)) -> c_15(choice^#(clause), guess^#(cnf)) , verify^#(cons(l, ls)) -> c_17(if^#(member(negate(l), ls), false(), verify(ls))) , sat^#(cnf) -> c_18(satck^#(cnf, guess(cnf))) , satck^#(cnf, assign) -> c_19(if^#(verify(assign), assign, unsat())) } Strict Trs: { if(true(), t, e) -> t , if(false(), t, e) -> e , member(x, nil()) -> false() , member(x, cons(y, ys)) -> if(eq(x, y), true(), member(x, ys)) , eq(nil(), nil()) -> true() , eq(O(x), 0(y)) -> eq(x, y) , eq(0(x), 1(y)) -> false() , eq(1(x), 0(y)) -> false() , eq(1(x), 1(y)) -> eq(x, y) , negate(0(x)) -> 1(x) , negate(1(x)) -> 0(x) , choice(cons(x, xs)) -> x , choice(cons(x, xs)) -> choice(xs) , guess(nil()) -> nil() , guess(cons(clause, cnf)) -> cons(choice(clause), guess(cnf)) , verify(nil()) -> true() , verify(cons(l, ls)) -> if(member(negate(l), ls), false(), verify(ls)) , sat(cnf) -> satck(cnf, guess(cnf)) , satck(cnf, assign) -> if(verify(assign), assign, unsat()) } Weak DPs: { member^#(x, nil()) -> c_3() , eq^#(nil(), nil()) -> c_5() , eq^#(0(x), 1(y)) -> c_7() , eq^#(1(x), 0(y)) -> c_8() , guess^#(nil()) -> c_14() , verify^#(nil()) -> c_16() } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. Arrrr..