*** 1 Progress [(O(1),O(n^1))] ***
Considered Problem:
Strict DP Rules:
Strict TRS Rules:
__(X1,mark(X2)) -> mark(__(X1,X2))
__(mark(X1),X2) -> mark(__(X1,X2))
__(ok(X1),ok(X2)) -> ok(__(X1,X2))
active(__(X,nil())) -> mark(X)
active(__(X1,X2)) -> __(X1,active(X2))
active(__(X1,X2)) -> __(active(X1),X2)
active(__(__(X,Y),Z)) -> mark(__(X,__(Y,Z)))
active(__(nil(),X)) -> mark(X)
active(and(X1,X2)) -> and(active(X1),X2)
active(and(tt(),X)) -> mark(X)
active(isNePal(X)) -> isNePal(active(X))
active(isNePal(__(I,__(P,I)))) -> mark(tt())
and(mark(X1),X2) -> mark(and(X1,X2))
and(ok(X1),ok(X2)) -> ok(and(X1,X2))
isNePal(mark(X)) -> mark(isNePal(X))
isNePal(ok(X)) -> ok(isNePal(X))
proper(__(X1,X2)) -> __(proper(X1),proper(X2))
proper(and(X1,X2)) -> and(proper(X1),proper(X2))
proper(isNePal(X)) -> isNePal(proper(X))
proper(nil()) -> ok(nil())
proper(tt()) -> ok(tt())
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))
Weak DP Rules:
Weak TRS Rules:
Signature:
{__/2,active/1,and/2,isNePal/1,proper/1,top/1} / {mark/1,nil/0,ok/1,tt/0}
Obligation:
Full
basic terms: {__,active,and,isNePal,proper,top}/{mark,nil,ok,tt}
Applied Processor:
Bounds {initialAutomaton = perSymbol, enrichment = match}
Proof:
The problem is match-bounded by 2.
The enriched problem is compatible with follwoing automaton.
___0(5,5) -> 1
___0(5,6) -> 1
___0(5,7) -> 1
___0(5,10) -> 1
___0(6,5) -> 1
___0(6,6) -> 1
___0(6,7) -> 1
___0(6,10) -> 1
___0(7,5) -> 1
___0(7,6) -> 1
___0(7,7) -> 1
___0(7,10) -> 1
___0(10,5) -> 1
___0(10,6) -> 1
___0(10,7) -> 1
___0(10,10) -> 1
___1(5,5) -> 11
___1(5,6) -> 11
___1(5,7) -> 11
___1(5,10) -> 11
___1(6,5) -> 11
___1(6,6) -> 11
___1(6,7) -> 11
___1(6,10) -> 11
___1(7,5) -> 11
___1(7,6) -> 11
___1(7,7) -> 11
___1(7,10) -> 11
___1(10,5) -> 11
___1(10,6) -> 11
___1(10,7) -> 11
___1(10,10) -> 11
active_0(5) -> 2
active_0(6) -> 2
active_0(7) -> 2
active_0(10) -> 2
active_1(5) -> 15
active_1(6) -> 15
active_1(7) -> 15
active_1(10) -> 15
active_2(14) -> 16
and_0(5,5) -> 3
and_0(5,6) -> 3
and_0(5,7) -> 3
and_0(5,10) -> 3
and_0(6,5) -> 3
and_0(6,6) -> 3
and_0(6,7) -> 3
and_0(6,10) -> 3
and_0(7,5) -> 3
and_0(7,6) -> 3
and_0(7,7) -> 3
and_0(7,10) -> 3
and_0(10,5) -> 3
and_0(10,6) -> 3
and_0(10,7) -> 3
and_0(10,10) -> 3
and_1(5,5) -> 12
and_1(5,6) -> 12
and_1(5,7) -> 12
and_1(5,10) -> 12
and_1(6,5) -> 12
and_1(6,6) -> 12
and_1(6,7) -> 12
and_1(6,10) -> 12
and_1(7,5) -> 12
and_1(7,6) -> 12
and_1(7,7) -> 12
and_1(7,10) -> 12
and_1(10,5) -> 12
and_1(10,6) -> 12
and_1(10,7) -> 12
and_1(10,10) -> 12
isNePal_0(5) -> 4
isNePal_0(6) -> 4
isNePal_0(7) -> 4
isNePal_0(10) -> 4
isNePal_1(5) -> 13
isNePal_1(6) -> 13
isNePal_1(7) -> 13
isNePal_1(10) -> 13
mark_0(5) -> 5
mark_0(6) -> 5
mark_0(7) -> 5
mark_0(10) -> 5
mark_1(11) -> 1
mark_1(11) -> 11
mark_1(12) -> 3
mark_1(12) -> 12
mark_1(13) -> 4
mark_1(13) -> 13
nil_0() -> 6
nil_1() -> 14
ok_0(5) -> 7
ok_0(6) -> 7
ok_0(7) -> 7
ok_0(10) -> 7
ok_1(11) -> 1
ok_1(11) -> 11
ok_1(12) -> 3
ok_1(12) -> 12
ok_1(13) -> 4
ok_1(13) -> 13
ok_1(14) -> 8
ok_1(14) -> 15
proper_0(5) -> 8
proper_0(6) -> 8
proper_0(7) -> 8
proper_0(10) -> 8
proper_1(5) -> 15
proper_1(6) -> 15
proper_1(7) -> 15
proper_1(10) -> 15
top_0(5) -> 9
top_0(6) -> 9
top_0(7) -> 9
top_0(10) -> 9
top_1(15) -> 9
top_2(16) -> 9
tt_0() -> 10
tt_1() -> 14
*** 1.1 Progress [(O(1),O(1))] ***
Considered Problem:
Strict DP Rules:
Strict TRS Rules:
Weak DP Rules:
Weak TRS Rules:
__(X1,mark(X2)) -> mark(__(X1,X2))
__(mark(X1),X2) -> mark(__(X1,X2))
__(ok(X1),ok(X2)) -> ok(__(X1,X2))
active(__(X,nil())) -> mark(X)
active(__(X1,X2)) -> __(X1,active(X2))
active(__(X1,X2)) -> __(active(X1),X2)
active(__(__(X,Y),Z)) -> mark(__(X,__(Y,Z)))
active(__(nil(),X)) -> mark(X)
active(and(X1,X2)) -> and(active(X1),X2)
active(and(tt(),X)) -> mark(X)
active(isNePal(X)) -> isNePal(active(X))
active(isNePal(__(I,__(P,I)))) -> mark(tt())
and(mark(X1),X2) -> mark(and(X1,X2))
and(ok(X1),ok(X2)) -> ok(and(X1,X2))
isNePal(mark(X)) -> mark(isNePal(X))
isNePal(ok(X)) -> ok(isNePal(X))
proper(__(X1,X2)) -> __(proper(X1),proper(X2))
proper(and(X1,X2)) -> and(proper(X1),proper(X2))
proper(isNePal(X)) -> isNePal(proper(X))
proper(nil()) -> ok(nil())
proper(tt()) -> ok(tt())
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))
Signature:
{__/2,active/1,and/2,isNePal/1,proper/1,top/1} / {mark/1,nil/0,ok/1,tt/0}
Obligation:
Full
basic terms: {__,active,and,isNePal,proper,top}/{mark,nil,ok,tt}
Applied Processor:
EmptyProcessor
Proof:
The problem is already closed. The intended complexity is O(1).