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:

append(Cons(x, xs), ys) → Cons(x, append(xs, ys))
append(Nil, ys) → ys
goal(x, y) → append(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:

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

Tuples:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:

append, goal

Defined Pair Symbols:

APPEND, GOAL

Compound Symbols:

c, c1, c2

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

GOAL(z0, z1) → c2(APPEND(z0, z1))

Removed 1 trailing nodes:

APPEND(Nil, z0) → c1

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))

S tuples:

APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))

K tuples:none
Defined Rule Symbols:

append, goal

Defined Pair Symbols:

APPEND

Compound Symbols:

c

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))

S tuples:

APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

APPEND

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.

APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))

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

APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))

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

POL(APPEND(x₁, x₂)) = [5]x₁   
POL(Cons(x₁, x₂)) = [1] + x₂   
POL(c(x₁)) = x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))

S tuples:none
K tuples:

APPEND(Cons(z0, z1), z2) → c(APPEND(z1, z2))

Defined Rule Symbols:none

Defined Pair Symbols:

APPEND

Compound Symbols:

c

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

The set S is empty

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