```* 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))
```