### (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(X) → if(X, c, n__f(true))
if(true, X, Y) → X
if(false, X, Y) → activate(Y)
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

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(X) → if(X, c, n__f(true))
if(true, X, Y) → X
if(false, X, Y) → activate(Y)
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

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 3.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2, 3]
transitions:
c0() → 0
n__f0(0) → 0
true0() → 0
false0() → 0
f0(0) → 1
if0(0, 0, 0) → 2
activate0(0) → 3
c1() → 4
true1() → 6
n__f1(6) → 5
if1(0, 4, 5) → 1
activate1(0) → 2
n__f1(0) → 1
f1(0) → 3
c2() → 7
true2() → 9
n__f2(9) → 8
if2(0, 7, 8) → 3
activate1(5) → 1
n__f2(0) → 3
f1(0) → 2
if2(0, 7, 8) → 2
n__f2(0) → 2
f2(6) → 1
activate1(8) → 3
activate1(8) → 2
f2(9) → 3
c3() → 10
true3() → 12
n__f3(12) → 11
if3(6, 10, 11) → 1
n__f3(6) → 1
f2(9) → 2
if3(9, 10, 11) → 3
n__f3(9) → 3
if3(9, 10, 11) → 2
n__f3(9) → 2
0 → 2
0 → 3
4 → 1
5 → 1
7 → 3
7 → 2
8 → 3
8 → 2
10 → 1
10 → 3
10 → 2