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 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 41 ms)
↳8 CdtProblem
↳9 SIsEmptyProof (⇔, 0 ms)
↳10 BOUNDS(O(1), O(1))

(0) Obligation:

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

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(c) → a__c
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(b, z0, c) → a__f(z0, a__c, z0)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__c → b
a__c → c
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2)
mark(c) → a__c
mark(b) → b
Tuples:

A__F(b, z0, c) → c1(A__F(z0, a__c, z0), A__C)
MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1))
MARK(c) → c6(A__C)
S tuples:

A__F(b, z0, c) → c1(A__F(z0, a__c, z0), A__C)
MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1))
MARK(c) → c6(A__C)
K tuples:none
Defined Rule Symbols:

a__f, a__c, mark

Defined Pair Symbols:

A__F, MARK

Compound Symbols:

c1, c5, c6

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

Removed 2 trailing nodes:

A__F(b, z0, c) → c1(A__F(z0, a__c, z0), A__C)
MARK(c) → c6(A__C)

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(b, z0, c) → a__f(z0, a__c, z0)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__c → b
a__c → c
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2)
mark(c) → a__c
mark(b) → b
Tuples:

MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1))
S tuples:

MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1))
K tuples:none
Defined Rule Symbols:

a__f, a__c, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c5

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

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(b, z0, c) → a__f(z0, a__c, z0)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__c → b
a__c → c
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2)
mark(c) → a__c
mark(b) → b
Tuples:

MARK(f(z0, z1, z2)) → c5(MARK(z1))
S tuples:

MARK(f(z0, z1, z2)) → c5(MARK(z1))
K tuples:none
Defined Rule Symbols:

a__f, a__c, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c5

(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, z1, z2)) → c5(MARK(z1))
We considered the (Usable) Rules:none
And the Tuples:

MARK(f(z0, z1, z2)) → c5(MARK(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(MARK(x1)) = [5]x1   
POL(c5(x1)) = x1   
POL(f(x1, x2, x3)) = [1] + x2   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(b, z0, c) → a__f(z0, a__c, z0)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__c → b
a__c → c
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2)
mark(c) → a__c
mark(b) → b
Tuples:

MARK(f(z0, z1, z2)) → c5(MARK(z1))
S tuples:none
K tuples:

MARK(f(z0, z1, z2)) → c5(MARK(z1))
Defined Rule Symbols:

a__f, a__c, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c5

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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