*** 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).