The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).

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

(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(n^2)) 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)) = [2]x₃ + [2]x₁·x₃ + [2]x₁²   
POL(c3(x₁)) = x₁   
POL(cons(x₁, x₂)) = 0   
POL(f(x₁)) = 0   
POL(nil) = 0   
POL(s(x₁)) = [2] + 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

(12) BOUNDS(1, 1)