### (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 Cpx (relative) TRS 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(empty, z0) → c
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(empty, z0) → c
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:

c, c1, c2

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

Removed 1 trailing nodes:

F(empty, z0) → c

### (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:

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

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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:none

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

### (7) 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.

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

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
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:none

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

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

The set S is empty