*** 1 Progress [(O(1),O(n^1))] ***
Considered Problem:
Strict DP Rules:
Strict TRS Rules:
add0(x,Nil()) -> x
add0(x',Cons(x,xs)) -> add0(Cons(Cons(Nil(),Nil()),x'),xs)
goal(x,y) -> add0(x,y)
notEmpty(Cons(x,xs)) -> True()
notEmpty(Nil()) -> False()
Weak DP Rules:
Weak TRS Rules:
Signature:
{add0/2,goal/2,notEmpty/1} / {Cons/2,False/0,Nil/0,True/0}
Obligation:
Innermost
basic terms: {add0,goal,notEmpty}/{Cons,False,Nil,True}
Applied Processor:
Bounds {initialAutomaton = perSymbol, enrichment = match}
Proof:
The problem is match-bounded by 1.
The enriched problem is compatible with follwoing automaton.
Cons_0(1,1) -> 1
Cons_0(1,1) -> 5
Cons_0(1,1) -> 6
Cons_0(1,2) -> 1
Cons_0(1,2) -> 5
Cons_0(1,2) -> 6
Cons_0(1,3) -> 1
Cons_0(1,3) -> 5
Cons_0(1,3) -> 6
Cons_0(1,4) -> 1
Cons_0(1,4) -> 5
Cons_0(1,4) -> 6
Cons_0(2,1) -> 1
Cons_0(2,1) -> 5
Cons_0(2,1) -> 6
Cons_0(2,2) -> 1
Cons_0(2,2) -> 5
Cons_0(2,2) -> 6
Cons_0(2,3) -> 1
Cons_0(2,3) -> 5
Cons_0(2,3) -> 6
Cons_0(2,4) -> 1
Cons_0(2,4) -> 5
Cons_0(2,4) -> 6
Cons_0(3,1) -> 1
Cons_0(3,1) -> 5
Cons_0(3,1) -> 6
Cons_0(3,2) -> 1
Cons_0(3,2) -> 5
Cons_0(3,2) -> 6
Cons_0(3,3) -> 1
Cons_0(3,3) -> 5
Cons_0(3,3) -> 6
Cons_0(3,4) -> 1
Cons_0(3,4) -> 5
Cons_0(3,4) -> 6
Cons_0(4,1) -> 1
Cons_0(4,1) -> 5
Cons_0(4,1) -> 6
Cons_0(4,2) -> 1
Cons_0(4,2) -> 5
Cons_0(4,2) -> 6
Cons_0(4,3) -> 1
Cons_0(4,3) -> 5
Cons_0(4,3) -> 6
Cons_0(4,4) -> 1
Cons_0(4,4) -> 5
Cons_0(4,4) -> 6
Cons_1(9,1) -> 5
Cons_1(9,1) -> 6
Cons_1(9,1) -> 8
Cons_1(9,2) -> 5
Cons_1(9,2) -> 6
Cons_1(9,2) -> 8
Cons_1(9,3) -> 5
Cons_1(9,3) -> 6
Cons_1(9,3) -> 8
Cons_1(9,4) -> 5
Cons_1(9,4) -> 6
Cons_1(9,4) -> 8
Cons_1(9,8) -> 5
Cons_1(9,8) -> 6
Cons_1(9,8) -> 8
Cons_1(10,11) -> 9
False_0() -> 2
False_0() -> 5
False_0() -> 6
False_1() -> 7
Nil_0() -> 3
Nil_0() -> 5
Nil_0() -> 6
Nil_1() -> 10
Nil_1() -> 11
True_0() -> 4
True_0() -> 5
True_0() -> 6
True_1() -> 7
add0_0(1,1) -> 5
add0_0(1,2) -> 5
add0_0(1,3) -> 5
add0_0(1,4) -> 5
add0_0(2,1) -> 5
add0_0(2,2) -> 5
add0_0(2,3) -> 5
add0_0(2,4) -> 5
add0_0(3,1) -> 5
add0_0(3,2) -> 5
add0_0(3,3) -> 5
add0_0(3,4) -> 5
add0_0(4,1) -> 5
add0_0(4,2) -> 5
add0_0(4,3) -> 5
add0_0(4,4) -> 5
add0_1(1,1) -> 6
add0_1(1,2) -> 6
add0_1(1,3) -> 6
add0_1(1,4) -> 6
add0_1(2,1) -> 6
add0_1(2,2) -> 6
add0_1(2,3) -> 6
add0_1(2,4) -> 6
add0_1(3,1) -> 6
add0_1(3,2) -> 6
add0_1(3,3) -> 6
add0_1(3,4) -> 6
add0_1(4,1) -> 6
add0_1(4,2) -> 6
add0_1(4,3) -> 6
add0_1(4,4) -> 6
add0_1(8,1) -> 5
add0_1(8,1) -> 6
add0_1(8,2) -> 5
add0_1(8,2) -> 6
add0_1(8,3) -> 5
add0_1(8,3) -> 6
add0_1(8,4) -> 5
add0_1(8,4) -> 6
goal_0(1,1) -> 6
goal_0(1,2) -> 6
goal_0(1,3) -> 6
goal_0(1,4) -> 6
goal_0(2,1) -> 6
goal_0(2,2) -> 6
goal_0(2,3) -> 6
goal_0(2,4) -> 6
goal_0(3,1) -> 6
goal_0(3,2) -> 6
goal_0(3,3) -> 6
goal_0(3,4) -> 6
goal_0(4,1) -> 6
goal_0(4,2) -> 6
goal_0(4,3) -> 6
goal_0(4,4) -> 6
notEmpty_0(1) -> 7
notEmpty_0(2) -> 7
notEmpty_0(3) -> 7
notEmpty_0(4) -> 7
1 -> 5
1 -> 6
2 -> 5
2 -> 6
3 -> 5
3 -> 6
4 -> 5
4 -> 6
8 -> 5
8 -> 6
*** 1.1 Progress [(O(1),O(1))] ***
Considered Problem:
Strict DP Rules:
Strict TRS Rules:
Weak DP Rules:
Weak TRS Rules:
add0(x,Nil()) -> x
add0(x',Cons(x,xs)) -> add0(Cons(Cons(Nil(),Nil()),x'),xs)
goal(x,y) -> add0(x,y)
notEmpty(Cons(x,xs)) -> True()
notEmpty(Nil()) -> False()
Signature:
{add0/2,goal/2,notEmpty/1} / {Cons/2,False/0,Nil/0,True/0}
Obligation:
Innermost
basic terms: {add0,goal,notEmpty}/{Cons,False,Nil,True}
Applied Processor:
EmptyProcessor
Proof:
The problem is already closed. The intended complexity is O(1).