The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

anchored(Cons(x, xs), y) → anchored(xs, Cons(Cons(Nil, Nil), y))
anchored(Nil, y) → y
goal(x, y) → anchored(x, y)

Rewrite Strategy: INNERMOST

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

anchored(Cons(z0, z1), z2) → anchored(z1, Cons(Cons(Nil, Nil), z2))
anchored(Nil, z0) → z0
goal(z0, z1) → anchored(z0, z1)

Tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))
ANCHORED(Nil, z0) → c1
GOAL(z0, z1) → c2(ANCHORED(z0, z1))

S tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))
ANCHORED(Nil, z0) → c1
GOAL(z0, z1) → c2(ANCHORED(z0, z1))

K tuples:none
Defined Rule Symbols:

anchored, goal

Defined Pair Symbols:

ANCHORED, GOAL

Compound Symbols:

c, c1, c2

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

GOAL(z0, z1) → c2(ANCHORED(z0, z1))

Removed 1 trailing nodes:

ANCHORED(Nil, z0) → c1

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

anchored(Cons(z0, z1), z2) → anchored(z1, Cons(Cons(Nil, Nil), z2))
anchored(Nil, z0) → z0
goal(z0, z1) → anchored(z0, z1)

Tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

S tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

K tuples:none
Defined Rule Symbols:

anchored, goal

Defined Pair Symbols:

ANCHORED

Compound Symbols:

c

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

anchored(Cons(z0, z1), z2) → anchored(z1, Cons(Cons(Nil, Nil), z2))
anchored(Nil, z0) → z0
goal(z0, z1) → anchored(z0, z1)

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

S tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

ANCHORED

Compound Symbols:

c

(7) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

We considered the (Usable) Rules:none
And the Tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ANCHORED(x₁, x₂)) = [4]x₁ + x₂   
POL(Cons(x₁, x₂)) = [5] + x₂   
POL(Nil) = [1]   
POL(c(x₁)) = x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

S tuples:none
K tuples:

ANCHORED(Cons(z0, z1), z2) → c(ANCHORED(z1, Cons(Cons(Nil, Nil), z2)))

Defined Rule Symbols:none

Defined Pair Symbols:

ANCHORED

Compound Symbols:

c

(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(10) BOUNDS(O(1), O(1))