### (0) Obligation:

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

f(cons(nil, y)) → y
f(cons(f(cons(nil, y)), z)) → copy(n, y, z)
copy(0, y, z) → f(z)
copy(s(x), y, z) → copy(x, y, cons(f(y), z))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(cons(nil, z0)) → z0
f(cons(f(cons(nil, z0)), z1)) → copy(n, z0, z1)
copy(0, z0, z1) → f(z1)
copy(s(z0), z1, z2) → copy(z0, z1, cons(f(z1), z2))
Tuples:

F(cons(nil, z0)) → c
F(cons(f(cons(nil, z0)), z1)) → c1(COPY(n, z0, z1))
COPY(0, z0, z1) → c2(F(z1))
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)), F(z1))
S tuples:

F(cons(nil, z0)) → c
F(cons(f(cons(nil, z0)), z1)) → c1(COPY(n, z0, z1))
COPY(0, z0, z1) → c2(F(z1))
COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)), F(z1))
K tuples:none
Defined Rule Symbols:

f, copy

Defined Pair Symbols:

F, COPY

Compound Symbols:

c, c1, c2, c3

### (3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 3 trailing nodes:

F(cons(nil, z0)) → c
F(cons(f(cons(nil, z0)), z1)) → c1(COPY(n, z0, z1))
COPY(0, z0, z1) → c2(F(z1))

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(cons(nil, z0)) → z0
f(cons(f(cons(nil, z0)), z1)) → copy(n, z0, z1)
copy(0, z0, z1) → f(z1)
copy(s(z0), z1, z2) → copy(z0, z1, cons(f(z1), z2))
Tuples:

COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)), F(z1))
S tuples:

COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)), F(z1))
K tuples:none
Defined Rule Symbols:

f, copy

Defined Pair Symbols:

COPY

Compound Symbols:

c3

### (5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(cons(nil, z0)) → z0
f(cons(f(cons(nil, z0)), z1)) → copy(n, z0, z1)
copy(0, z0, z1) → f(z1)
copy(s(z0), z1, z2) → copy(z0, z1, cons(f(z1), z2))
Tuples:

COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
S tuples:

COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
K tuples:none
Defined Rule Symbols:

f, copy

Defined Pair Symbols:

COPY

Compound Symbols:

c3

### (7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(cons(f(cons(nil, z0)), z1)) → copy(n, z0, z1)
copy(0, z0, z1) → f(z1)
copy(s(z0), z1, z2) → copy(z0, z1, cons(f(z1), z2))

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(cons(nil, z0)) → z0
Tuples:

COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
S tuples:

COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

COPY

Compound Symbols:

c3

### (9) 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.

COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
We considered the (Usable) Rules:none
And the Tuples:

COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(COPY(x1, x2, x3)) = [4]x1 + [4]x3
POL(c3(x1)) = x1
POL(cons(x1, x2)) = [3]
POL(f(x1)) = 0
POL(nil) = 0
POL(s(x1)) = [4] + x1

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(cons(nil, z0)) → z0
Tuples:

COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
S tuples:none
K tuples:

COPY(s(z0), z1, z2) → c3(COPY(z0, z1, cons(f(z1), z2)))
Defined Rule Symbols:

f

Defined Pair Symbols:

COPY

Compound Symbols:

c3

### (11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty