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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsProof (⇔, 16 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 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳8 CdtProblem
        ↳9 CdtUsableRulesProof (⇔, 0 ms)
        ↳10 CdtProblem
        ↳11 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 24 ms)
        ↳12 CdtProblem
        ↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳14 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)) → c(f(g(f(a))))
mark(f(X)) → a__f(mark(X))
mark(a) → a
mark(c(X)) → c(X)
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|1, c_1|1]
1→3[c_1|1]
1→7[a__f_1|1, f_1|2]
1→8[g_1|1]
1→9[c_1|2]
2→2[f_1|0, a|0, c_1|0, g_1|0]
3→4[f_1|1]
4→5[g_1|1]
5→6[f_1|1]
6→2[a|1]
7→2[mark_1|1, a|1, c_1|1]
7→7[a__f_1|1, f_1|2]
7→8[g_1|1]
7→9[c_1|2]
8→2[mark_1|1, a|1, c_1|1]
8→7[a__f_1|1, f_1|2]
8→8[g_1|1]
8→9[c_1|2]
9→10[f_1|2]
10→11[g_1|2]
11→12[f_1|2]
12→2[a|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)) → c(f(g(f(a))))
a__f(z0) → f(z0)
mark(f(z0)) → a__f(mark(z0))
mark(a) → a
mark(c(z0)) → c(z0)
mark(g(z0)) → g(mark(z0))
Tuples:

A__F(f(a)) → c1
A__F(z0) → c2
MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0))
MARK(a) → c4
MARK(c(z0)) → c5
MARK(g(z0)) → c6(MARK(z0))
S tuples:

A__F(f(a)) → c1
A__F(z0) → c2
MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0))
MARK(a) → c4
MARK(c(z0)) → c5
MARK(g(z0)) → c6(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

A__F, MARK

Compound Symbols:

c1, c2, c3, c4, c5, c6

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

Removed 4 trailing nodes:

MARK(a) → c4
A__F(f(a)) → c1
MARK(c(z0)) → c5
A__F(z0) → c2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0))
MARK(g(z0)) → c6(MARK(z0))
S tuples:

MARK(f(z0)) → c3(A__F(mark(z0)), MARK(z0))
MARK(g(z0)) → c6(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c3, c6

(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

MARK(g(z0)) → c6(MARK(z0))
MARK(f(z0)) → c3(MARK(z0))
S tuples:

MARK(g(z0)) → c6(MARK(z0))
MARK(f(z0)) → c3(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c6, c3

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MARK(g(z0)) → c6(MARK(z0))
MARK(f(z0)) → c3(MARK(z0))
S tuples:

MARK(g(z0)) → c6(MARK(z0))
MARK(f(z0)) → c3(MARK(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

MARK

Compound Symbols:

c6, c3

(11) 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)) → c6(MARK(z0))
MARK(f(z0)) → c3(MARK(z0))
We considered the (Usable) Rules:none
And the Tuples:

MARK(g(z0)) → c6(MARK(z0))
MARK(f(z0)) → c3(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(MARK(x1)) = [2]x1   
POL(c3(x1)) = x1   
POL(c6(x1)) = x1   
POL(f(x1)) = [1] + x1   
POL(g(x1)) = [1] + x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MARK(g(z0)) → c6(MARK(z0))
MARK(f(z0)) → c3(MARK(z0))
S tuples:none
K tuples:

MARK(g(z0)) → c6(MARK(z0))
MARK(f(z0)) → c3(MARK(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

MARK

Compound Symbols:

c6, c3

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

The set S is empty

(14) BOUNDS(1, 1)