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

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳4 CdtProblem
↳5 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 CdtProblem
↳7 CdtUsableRulesProof (⇔, 0 ms)
↳8 CdtProblem
↳9 CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))), 30 ms)
↳10 CdtProblem
↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳12 BOUNDS(O(1), O(1))

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

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

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) 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(mark(z0))
mark(a) → a
mark(g(z0)) → g(z0)
Tuples:

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

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

a__f, mark

Defined Pair Symbols:

A__F, MARK

Compound Symbols:

c, c1, c2, c3, c4

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

Removed 4 trailing nodes:

A__F(f(a)) → c(A__F(g(f(a))))
A__F(z0) → c1
MARK(a) → c3
MARK(g(z0)) → c4

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

MARK(f(z0)) → c2(A__F(mark(z0)), MARK(z0))
S tuples:

MARK(f(z0)) → c2(A__F(mark(z0)), MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c2

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

Removed 1 trailing tuple parts

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

MARK(f(z0)) → c2(MARK(z0))
S tuples:

MARK(f(z0)) → c2(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c2

(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(mark(z0))
mark(a) → a
mark(g(z0)) → g(z0)

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MARK(f(z0)) → c2(MARK(z0))
S tuples:

MARK(f(z0)) → c2(MARK(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

MARK

Compound Symbols:

c2

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

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

MARK(f(z0)) → c2(MARK(z0))
We considered the (Usable) Rules:none
And the Tuples:

MARK(f(z0)) → c2(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(MARK(x1)) = [5]x1   
POL(c2(x1)) = x1   
POL(f(x1)) = [1] + x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MARK(f(z0)) → c2(MARK(z0))
S tuples:none
K tuples:

MARK(f(z0)) → c2(MARK(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

MARK

Compound Symbols:

c2

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

The set S is empty

(12) BOUNDS(O(1), O(1))