* Step 1: Sum WORST_CASE(Omega(n^1),O(n^1)) + Considered Problem: - Strict TRS: dfsAcc#3(Leaf(x8),x16) -> Cons(x8,x16) dfsAcc#3(Node(x6,x4),x2) -> dfsAcc#3(x4,dfsAcc#3(x6,x2)) main(x1) -> revApp#2(dfsAcc#3(x1,Nil()),Nil()) revApp#2(Cons(x6,x4),x2) -> revApp#2(x4,Cons(x6,x2)) revApp#2(Nil(),x16) -> x16 - Signature: {dfsAcc#3/2,main/1,revApp#2/2} / {Cons/2,Leaf/1,Nil/0,Node/2} - Obligation: innermost runtime complexity wrt. defined symbols {dfsAcc#3,main,revApp#2} and constructors {Cons,Leaf,Nil ,Node} + Applied Processor: Sum {left = someStrategy, right = someStrategy} + Details: () ** Step 1.a:1: DecreasingLoops WORST_CASE(Omega(n^1),?) + Considered Problem: - Strict TRS: dfsAcc#3(Leaf(x8),x16) -> Cons(x8,x16) dfsAcc#3(Node(x6,x4),x2) -> dfsAcc#3(x4,dfsAcc#3(x6,x2)) main(x1) -> revApp#2(dfsAcc#3(x1,Nil()),Nil()) revApp#2(Cons(x6,x4),x2) -> revApp#2(x4,Cons(x6,x2)) revApp#2(Nil(),x16) -> x16 - Signature: {dfsAcc#3/2,main/1,revApp#2/2} / {Cons/2,Leaf/1,Nil/0,Node/2} - Obligation: innermost runtime complexity wrt. defined symbols {dfsAcc#3,main,revApp#2} and constructors {Cons,Leaf,Nil ,Node} + Applied Processor: DecreasingLoops {bound = AnyLoop, narrow = 10} + Details: The system has following decreasing Loops: dfsAcc#3(y,z){y -> Node(x,y)} = dfsAcc#3(Node(x,y),z) ->^+ dfsAcc#3(y,dfsAcc#3(x,z)) = C[dfsAcc#3(y,dfsAcc#3(x,z)) = dfsAcc#3(y,z){z -> dfsAcc#3(x,z)}] ** Step 1.b:1: Bounds WORST_CASE(?,O(n^1)) + Considered Problem: - Strict TRS: dfsAcc#3(Leaf(x8),x16) -> Cons(x8,x16) dfsAcc#3(Node(x6,x4),x2) -> dfsAcc#3(x4,dfsAcc#3(x6,x2)) main(x1) -> revApp#2(dfsAcc#3(x1,Nil()),Nil()) revApp#2(Cons(x6,x4),x2) -> revApp#2(x4,Cons(x6,x2)) revApp#2(Nil(),x16) -> x16 - Signature: {dfsAcc#3/2,main/1,revApp#2/2} / {Cons/2,Leaf/1,Nil/0,Node/2} - Obligation: innermost runtime complexity wrt. defined symbols {dfsAcc#3,main,revApp#2} and constructors {Cons,Leaf,Nil ,Node} + Applied Processor: Bounds {initialAutomaton = minimal, enrichment = match} + Details: The problem is match-bounded by 2. The enriched problem is compatible with follwoing automaton. Cons_0(2,2) -> 1 Cons_0(2,2) -> 2 Cons_1(2,1) -> 1 Cons_1(2,2) -> 1 Cons_1(2,2) -> 3 Cons_1(2,3) -> 1 Cons_1(2,3) -> 3 Cons_1(2,4) -> 4 Cons_1(2,6) -> 4 Cons_2(2,5) -> 1 Cons_2(2,5) -> 7 Cons_2(2,7) -> 1 Cons_2(2,7) -> 7 Leaf_0(2) -> 1 Leaf_0(2) -> 2 Nil_0() -> 1 Nil_0() -> 2 Nil_1() -> 5 Nil_1() -> 6 Node_0(2,2) -> 1 Node_0(2,2) -> 2 dfsAcc#3_0(2,2) -> 1 dfsAcc#3_1(2,2) -> 3 dfsAcc#3_1(2,3) -> 1 dfsAcc#3_1(2,3) -> 3 dfsAcc#3_1(2,4) -> 4 dfsAcc#3_1(2,6) -> 4 main_0(2) -> 1 revApp#2_0(2,2) -> 1 revApp#2_1(2,1) -> 1 revApp#2_1(4,5) -> 1 revApp#2_2(4,7) -> 1 revApp#2_2(6,7) -> 1 2 -> 1 7 -> 1 ** Step 1.b:2: EmptyProcessor WORST_CASE(?,O(1)) + Considered Problem: - Weak TRS: dfsAcc#3(Leaf(x8),x16) -> Cons(x8,x16) dfsAcc#3(Node(x6,x4),x2) -> dfsAcc#3(x4,dfsAcc#3(x6,x2)) main(x1) -> revApp#2(dfsAcc#3(x1,Nil()),Nil()) revApp#2(Cons(x6,x4),x2) -> revApp#2(x4,Cons(x6,x2)) revApp#2(Nil(),x16) -> x16 - Signature: {dfsAcc#3/2,main/1,revApp#2/2} / {Cons/2,Leaf/1,Nil/0,Node/2} - Obligation: innermost runtime complexity wrt. defined symbols {dfsAcc#3,main,revApp#2} and constructors {Cons,Leaf,Nil ,Node} + Applied Processor: EmptyProcessor + Details: The problem is already closed. The intended complexity is O(1). WORST_CASE(Omega(n^1),O(n^1))