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:

active(f(f(a))) → mark(f(g(f(a))))
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Rewrite Strategy: FULL

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

The following defined symbols can occur below the 0th argument of top: proper, active
The following defined symbols can occur below the 0th argument of proper: proper, active
The following defined symbols can occur below the 0th argument of active: proper, active

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
active(f(f(a))) → mark(f(g(f(a))))
active(g(X)) → g(active(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(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:

g(ok(X)) → ok(g(X))
top(ok(X)) → top(active(X))
f(ok(X)) → ok(f(X))
g(mark(X)) → mark(g(X))
top(mark(X)) → top(proper(X))
proper(a) → ok(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:

g(ok(X)) → ok(g(X))
top(ok(X)) → top(active(X))
f(ok(X)) → ok(f(X))
g(mark(X)) → mark(g(X))
top(mark(X)) → top(proper(X))
proper(a) → ok(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: 9
Accept states: [10]
Transitions:
9→10[g_1|0, top_1|0, f_1|0, proper_1|0]
9→11[ok_1|1]
9→12[mark_1|1]
9→13[top_1|1]
9→14[top_1|1]
9→15[ok_1|1]
9→16[ok_1|1]
9→17[top_1|2]
10→10[ok_1|0, active_1|0, mark_1|0, a|0]
11→10[g_1|1]
11→11[ok_1|1]
11→12[mark_1|1]
12→10[g_1|1]
12→11[ok_1|1]
12→12[mark_1|1]
13→10[active_1|1]
14→10[proper_1|1]
14→16[ok_1|1]
15→10[f_1|1]
15→15[ok_1|1]
16→10[a|1]
17→16[active_1|2]

(6) BOUNDS(1, n^1)