```* Step 1: Bounds WORST_CASE(?,O(n^1))
+ Considered Problem:
- Strict TRS:
cons(mark(X1),X2) -> mark(cons(X1,X2))
cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
fst(X1,mark(X2)) -> mark(fst(X1,X2))
fst(mark(X1),X2) -> mark(fst(X1,X2))
fst(ok(X1),ok(X2)) -> ok(fst(X1,X2))
len(mark(X)) -> mark(len(X))
len(ok(X)) -> ok(len(X))
proper(0()) -> ok(0())
proper(nil()) -> ok(nil())
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))
- Signature:
- Obligation:
innermost runtime complexity wrt. defined symbols {add,cons,from,fst,len,proper,s,top} and constructors {0
,active,mark,nil,ok}
+ Applied Processor:
Bounds {initialAutomaton = minimal, enrichment = match}
+ Details:
The problem is match-bounded by 2.
The enriched problem is compatible with follwoing automaton.
0_0() -> 2
0_1() -> 3
active_0(2) -> 2
active_1(2) -> 4
active_2(3) -> 5
cons_0(2,2) -> 1
cons_1(2,2) -> 3
from_0(2) -> 1
from_1(2) -> 3
fst_0(2,2) -> 1
fst_1(2,2) -> 3
len_0(2) -> 1
len_1(2) -> 3
mark_0(2) -> 2
mark_1(3) -> 1
mark_1(3) -> 3
nil_0() -> 2
nil_1() -> 3
ok_0(2) -> 2
ok_1(3) -> 1
ok_1(3) -> 3
ok_1(3) -> 4
proper_0(2) -> 1
proper_1(2) -> 4
s_0(2) -> 1
s_1(2) -> 3
top_0(2) -> 1
top_1(4) -> 1
top_2(5) -> 1
* Step 2: EmptyProcessor WORST_CASE(?,O(1))
+ Considered Problem:
- Weak TRS:
cons(mark(X1),X2) -> mark(cons(X1,X2))
cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
fst(X1,mark(X2)) -> mark(fst(X1,X2))
fst(mark(X1),X2) -> mark(fst(X1,X2))
fst(ok(X1),ok(X2)) -> ok(fst(X1,X2))
len(mark(X)) -> mark(len(X))
len(ok(X)) -> ok(len(X))
proper(0()) -> ok(0())
proper(nil()) -> ok(nil())
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))
- Signature: