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:

f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(empty, z0) → z0
f(cons(z0, z1), z2) → g(z1, z2, cons(z0, z1))
g(z0, z1, z2) → f(z0, cons(z1, z2))

Tuples:

F(cons(z0, z1), z2) → c1(G(z1, z2, cons(z0, z1)))
G(z0, z1, z2) → c2(F(z0, cons(z1, z2)))

S tuples:

F(cons(z0, z1), z2) → c1(G(z1, z2, cons(z0, z1)))
G(z0, z1, z2) → c2(F(z0, cons(z1, z2)))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

(3) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

F(cons(z0, z1), z2) → c1(G(z1, z2, cons(z0, z1)))
G(z0, z1, z2) → c2(F(z0, cons(z1, z2)))

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

F(cons(z0, z1), z2) → c1(G(z1, z2, cons(z0, z1)))
G(z0, z1, z2) → c2(F(z0, cons(z1, z2)))

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

POL(F(x₁, x₂)) = [2] + [4]x₁   
POL(G(x₁, x₂, x₃)) = [4] + [4]x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(cons(x₁, x₂)) = [4] + x₁ + x₂   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(empty, z0) → z0
f(cons(z0, z1), z2) → g(z1, z2, cons(z0, z1))
g(z0, z1, z2) → f(z0, cons(z1, z2))

Tuples:

F(cons(z0, z1), z2) → c1(G(z1, z2, cons(z0, z1)))
G(z0, z1, z2) → c2(F(z0, cons(z1, z2)))

S tuples:none
K tuples:

F(cons(z0, z1), z2) → c1(G(z1, z2, cons(z0, z1)))
G(z0, z1, z2) → c2(F(z0, cons(z1, z2)))

Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(6) BOUNDS(O(1), O(1))