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

c1, c2, c3

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

Removed 2 trailing nodes:

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) CdtPolyRedPairProof (UPPER BOUND (ADD(O(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)), F(z1))

We considered the (Usable) Rules:

f(cons(nil, z0)) → z0

And the Tuples:

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

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

POL(COPY(x₁, x₂, x₃)) = [3]x₁ + [2]x₃ + [2]x₁·x₂   
POL(F(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(cons(x₁, x₂)) = [2]   
POL(f(x₁)) = 0   
POL(nil) = 0   
POL(s(x₁)) = [3] + x₁   

(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)), F(z1))

S tuples:none
K tuples:

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

Defined Rule Symbols:

f, copy

Defined Pair Symbols:

COPY

Compound Symbols:

c3

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))