### (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)

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