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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsProof (⇔, 0 ms)
    ↳2 BOUNDS(1, n^1)
    ↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳4 CdtProblem
        ↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳6 CdtProblem
        ↳7 CdtUsableRulesProof (⇔, 0 ms)
        ↳8 CdtProblem
        ↳9 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 23 ms)
        ↳10 CdtProblem
        ↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳12 BOUNDS(1, 1)

(0) 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:

a__f(f(a)) → a__f(g(f(a)))
mark(f(X)) → a__f(X)
mark(a) → a
mark(g(X)) → g(mark(X))
a__f(X) → f(X)

Rewrite Strategy: INNERMOST

(1) 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: 1
Accept states: [2]
Transitions:
1→2[a__f_1|0, mark_1|0, f_1|1, a__f_1|1, a|1, f_1|2]
1→3[a__f_1|1, f_1|2]
1→6[g_1|1]
2→2[f_1|0, a|0, g_1|0]
3→4[g_1|1]
4→5[f_1|1]
5→2[a|1]
6→2[mark_1|1, a__f_1|1, a|1, f_1|2]
6→6[g_1|1]
6→3[a__f_1|1, f_1|2]

(2) BOUNDS(1, n^1)

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(f(a)) → a__f(g(f(a)))
a__f(z0) → f(z0)
mark(f(z0)) → a__f(z0)
mark(a) → a
mark(g(z0)) → g(mark(z0))
Tuples:

A__F(f(a)) → c(A__F(g(f(a))))
A__F(z0) → c1
MARK(f(z0)) → c2(A__F(z0))
MARK(a) → c3
MARK(g(z0)) → c4(MARK(z0))
S tuples:

A__F(f(a)) → c(A__F(g(f(a))))
A__F(z0) → c1
MARK(f(z0)) → c2(A__F(z0))
MARK(a) → c3
MARK(g(z0)) → c4(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

A__F, MARK

Compound Symbols:

c, c1, c2, c3, c4

(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 4 trailing nodes:

A__F(f(a)) → c(A__F(g(f(a))))
MARK(f(z0)) → c2(A__F(z0))
A__F(z0) → c1
MARK(a) → c3

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(f(a)) → a__f(g(f(a)))
a__f(z0) → f(z0)
mark(f(z0)) → a__f(z0)
mark(a) → a
mark(g(z0)) → g(mark(z0))
Tuples:

MARK(g(z0)) → c4(MARK(z0))
S tuples:

MARK(g(z0)) → c4(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c4

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

a__f(f(a)) → a__f(g(f(a)))
a__f(z0) → f(z0)
mark(f(z0)) → a__f(z0)
mark(a) → a
mark(g(z0)) → g(mark(z0))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MARK(g(z0)) → c4(MARK(z0))
S tuples:

MARK(g(z0)) → c4(MARK(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

MARK

Compound Symbols:

c4

(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

MARK(g(z0)) → c4(MARK(z0))
We considered the (Usable) Rules:none
And the Tuples:

MARK(g(z0)) → c4(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(MARK(x1)) = x1   
POL(c4(x1)) = x1   
POL(g(x1)) = [1] + x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MARK(g(z0)) → c4(MARK(z0))
S tuples:none
K tuples:

MARK(g(z0)) → c4(MARK(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

MARK

Compound Symbols:

c4

(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(12) BOUNDS(1, 1)