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) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(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(a) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(X)

Rewrite Strategy: INNERMOST

(3) 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 3.
The certificate found is represented by the following graph.

Start state: 3
Accept states: [4]
Transitions:
3→4[f_1|0, e_1|0, a|1, c_1|1, d_1|1, b|1, g_1|1, e_1|1, a|2, b|2, b|3]
3→5[f_1|1]
3→7[f_1|1]
3→9[f_1|1]
3→11[f_1|2]
4→4[a|0, c_1|0, d_1|0, b|0, g_1|0]
5→6[c_1|1]
6→4[a|1]
7→8[d_1|1]
8→4[a|1]
9→10[d_1|1]
10→4[b|1]
11→12[d_1|2]
12→4[b|2]

(4) BOUNDS(1, n^1)