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(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

S tuples:

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c3

(3) 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.

F(z0, empty) → c(G(z0, empty))

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

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

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

POL(F(x₁, x₂)) = [1] + x₁ + x₁·x₂   
POL(G(x₁, x₂)) = 0   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(cons(x₁, x₂)) = 0   
POL(empty) = [1]   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

S tuples:

F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

K tuples:

F(z0, empty) → c(G(z0, empty))

Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c3

(5) 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(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

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

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

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

POL(F(x₁, x₂)) = [4]x₁ + [5]x₂   
POL(G(x₁, x₂)) = [2] + [4]x₁   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(cons(x₁, x₂)) = [4] + x₁ + x₂   
POL(empty) = [5]   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

S tuples:none
K tuples:

F(z0, empty) → c(G(z0, empty))
F(z0, cons(z1, z2)) → c1(F(cons(z1, z0), z2))
G(cons(z0, z1), z2) → c3(G(z1, cons(z0, z2)))

Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c3

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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