### (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(0) → s(0)
f(s(0)) → s(s(0))
f(s(0)) → *(s(s(0)), f(0))
f(+(x, s(0))) → +(s(s(0)), f(x))
f(+(x, y)) → *(f(x), f(y))

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(0) → s(0)
f(s(0)) → s(s(0))
f(s(0)) → *(s(s(0)), f(0))
f(+(x, s(0))) → +(s(s(0)), f(x))
f(+(x, y)) → *(f(x), f(y))

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]
transitions:
00() → 0
s0(0) → 0
*0(0, 0) → 0
+0(0, 0) → 0
f0(0) → 1
01() → 2
s1(2) → 1
s1(2) → 3
s1(3) → 1
s1(3) → 4
01() → 6
f1(6) → 5
*1(4, 5) → 1
s1(3) → 7
f1(0) → 8
+1(7, 8) → 1
f1(0) → 9
f1(0) → 10
*1(9, 10) → 1
s1(2) → 8
s1(2) → 9
s1(2) → 10
02() → 11
s2(11) → 5
s1(3) → 8
s1(3) → 9
s1(3) → 10
*1(4, 5) → 8
*1(4, 5) → 9
*1(4, 5) → 10
+1(7, 8) → 8
+1(7, 8) → 9
+1(7, 8) → 10
*1(9, 10) → 8
*1(9, 10) → 9
*1(9, 10) → 10