### (0) Obligation:

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

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: FULL

### (1) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)

Converted rc-obligation to irc-obligation.

As the TRS does not nest defined symbols, we have rc = irc.

### (2) Obligation:

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

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

### (3) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 2.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2]
transitions:
empty0() → 0
cons0(0, 0) → 0
f0(0, 0) → 1
g0(0, 0) → 2
empty1() → 3
g1(0, 3) → 1
cons1(0, 0) → 4
f1(4, 0) → 1
cons1(0, 0) → 5
g1(0, 5) → 2
g1(4, 3) → 1
cons1(0, 4) → 4
cons1(0, 3) → 5
g1(0, 5) → 1
cons1(0, 5) → 5
cons2(0, 3) → 6
g2(0, 6) → 1
g2(4, 6) → 1
cons1(0, 6) → 5
cons2(0, 6) → 6
0 → 2
3 → 1
5 → 2
5 → 1
6 → 1