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

Converted Cpx (relative) TRS to CDT

(2) 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

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

Removed 4 trailing nodes:

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

(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:

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

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

Removed 1 trailing tuple parts

(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(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

(7) 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))

(8) 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

(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(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)) = [4]x₁   
POL(c3(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(f(x₁)) = [4] + x₁   
POL(g(x₁)) = [1] + x₁   

(10) 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

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

The set S is empty

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