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

(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) → g(h(a))
h(g(x)) → g(h(f(x)))
k(x, h(x), a) → h(x)
k(f(x), y, x) → f(x)

Rewrite Strategy: FULL

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

The following defined symbols can occur below the 0th argument of h: f, h
The following defined symbols can occur below the 0th argument of f: f, h

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
k(x, h(x), a) → h(x)
k(f(x), y, x) → f(x)

(2) 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:

h(g(x)) → g(h(f(x)))
f(a) → g(h(a))

Rewrite Strategy: FULL

(3) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)

Converted rc-obligation to irc-obligation.

As the TRS is a non-duplicating overlay system, we have rc = irc.

(4) 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:

h(g(x)) → g(h(f(x)))
f(a) → g(h(a))

Rewrite Strategy: INNERMOST

(5) CpxTrsMatchBoundsProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 2.
The certificate found is represented by the following graph.

Start state: 5
Accept states: [6]
Transitions:
5→6[h_1|0, f_1|0]
5→7[g_1|1]
5→9[g_1|1]
6→6[g_1|0, a|0]
7→8[h_1|1]
7→11[g_1|2]
8→6[f_1|1]
8→9[g_1|1]
9→10[h_1|1]
10→6[a|1]
11→12[h_1|2]
12→9[f_1|2]

(6) BOUNDS(1, n^1)