The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^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 CpxTRS 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))))
MARK(f(z0)) → c2(A__F(mark(z0)), MARK(z0))

S tuples:

A__F(f(a)) → c(A__F(g(f(a))))
MARK(f(z0)) → c2(A__F(mark(z0)), MARK(z0))

K tuples:none
Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

A__F, MARK

Compound Symbols:

c, c2

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

Removed 1 trailing nodes:

A__F(f(a)) → c(A__F(g(f(a))))

(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) CdtPolyRedPairProof (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   

(8) 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:none
K tuples:

MARK(f(z0)) → c2(MARK(z0))

Defined Rule Symbols:

a__f, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c2

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(10) BOUNDS(O(1), O(1))