* Step 1: Bounds WORST_CASE(?,O(n^1)) + Considered Problem: - Strict TRS: active(f(x)) -> f(active(x)) active(f(x)) -> mark(x) check(x) -> start(match(f(X()),x)) check(f(x)) -> f(check(x)) f(found(x)) -> found(f(x)) f(mark(x)) -> mark(f(x)) f(ok(x)) -> ok(f(x)) match(X(),x) -> proper(x) proper(c()) -> ok(c()) start(ok(x)) -> found(x) top(active(c())) -> top(mark(c())) top(found(x)) -> top(active(x)) top(mark(x)) -> top(check(x)) - Signature: {active/1,check/1,f/1,match/2,proper/1,start/1,top/1} / {X/0,c/0,found/1,mark/1,ok/1} - Obligation: runtime complexity wrt. defined symbols {active,check,f,match,proper,start,top} and constructors {X,c,found ,mark,ok} + Applied Processor: Bounds {initialAutomaton = minimal, enrichment = match} + Details: The problem is match-bounded by 4. The enriched problem is compatible with follwoing automaton. X_0() -> 2 X_1() -> 5 X_2() -> 10 X_3() -> 15 X_4() -> 20 active_0(2) -> 1 active_1(2) -> 7 active_2(2) -> 12 active_2(6) -> 11 c_0() -> 2 c_1() -> 6 c_2() -> 2 check_0(2) -> 1 check_1(2) -> 7 check_2(2) -> 12 check_2(6) -> 11 check_3(2) -> 16 f_0(2) -> 1 f_1(2) -> 6 f_1(5) -> 4 f_2(10) -> 9 f_2(12) -> 11 f_3(15) -> 14 f_4(20) -> 19 found_0(2) -> 2 found_1(2) -> 1 found_1(6) -> 1 found_1(6) -> 6 mark_0(2) -> 2 mark_1(6) -> 1 mark_1(6) -> 6 mark_2(2) -> 11 match_0(2,2) -> 1 match_1(4,2) -> 3 match_2(9,2) -> 8 match_3(14,2) -> 17 match_3(14,6) -> 13 match_4(19,2) -> 18 ok_0(2) -> 2 ok_1(6) -> 1 ok_1(6) -> 6 ok_2(2) -> 1 proper_0(2) -> 1 proper_1(2) -> 1 start_0(2) -> 1 start_1(3) -> 1 start_2(8) -> 7 start_3(13) -> 11 start_3(17) -> 12 start_4(18) -> 16 top_0(2) -> 1 top_1(6) -> 1 top_1(7) -> 1 top_2(11) -> 1 top_3(16) -> 1 * Step 2: EmptyProcessor WORST_CASE(?,O(1)) + Considered Problem: - Weak TRS: active(f(x)) -> f(active(x)) active(f(x)) -> mark(x) check(x) -> start(match(f(X()),x)) check(f(x)) -> f(check(x)) f(found(x)) -> found(f(x)) f(mark(x)) -> mark(f(x)) f(ok(x)) -> ok(f(x)) match(X(),x) -> proper(x) proper(c()) -> ok(c()) start(ok(x)) -> found(x) top(active(c())) -> top(mark(c())) top(found(x)) -> top(active(x)) top(mark(x)) -> top(check(x)) - Signature: {active/1,check/1,f/1,match/2,proper/1,start/1,top/1} / {X/0,c/0,found/1,mark/1,ok/1} - Obligation: runtime complexity wrt. defined symbols {active,check,f,match,proper,start,top} and constructors {X,c,found ,mark,ok} + Applied Processor: EmptyProcessor + Details: The problem is already closed. The intended complexity is O(1). WORST_CASE(?,O(n^1))