The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, 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(n^1)), 34 ms)
↳10 CdtProblem
↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID), 1 ms)
↳12 BOUNDS(1, 1)

(0) Obligation:

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

a__f(X, X) → a__f(a, b)
a__b → a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
a__b → b

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(z0, z0) → a__f(a, b)
a__f(z0, z1) → f(z0, z1)
a__b → a
a__b → b
mark(f(z0, z1)) → a__f(mark(z0), z1)
mark(b) → a__b
mark(a) → a
Tuples:

A__F(z0, z0) → c(A__F(a, b))
A__F(z0, z1) → c1
A__B → c2
A__B → c3
MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0))
MARK(b) → c5(A__B)
MARK(a) → c6
S tuples:

A__F(z0, z0) → c(A__F(a, b))
A__F(z0, z1) → c1
A__B → c2
A__B → c3
MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0))
MARK(b) → c5(A__B)
MARK(a) → c6
K tuples:none
Defined Rule Symbols:

a__f, a__b, mark

Defined Pair Symbols:

A__F, A__B, MARK

Compound Symbols:

c, c1, c2, c3, c4, c5, c6

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

Removed 6 trailing nodes:

MARK(a) → c6
A__F(z0, z0) → c(A__F(a, b))
A__B → c3
A__B → c2
MARK(b) → c5(A__B)
A__F(z0, z1) → c1

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0, z0) → a__f(a, b)
a__f(z0, z1) → f(z0, z1)
a__b → a
a__b → b
mark(f(z0, z1)) → a__f(mark(z0), z1)
mark(b) → a__b
mark(a) → a
Tuples:

MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0))
S tuples:

MARK(f(z0, z1)) → c4(A__F(mark(z0), z1), MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, a__b, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c4

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

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0, z0) → a__f(a, b)
a__f(z0, z1) → f(z0, z1)
a__b → a
a__b → b
mark(f(z0, z1)) → a__f(mark(z0), z1)
mark(b) → a__b
mark(a) → a
Tuples:

MARK(f(z0, z1)) → c4(MARK(z0))
S tuples:

MARK(f(z0, z1)) → c4(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, a__b, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c4

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

a__f(z0, z0) → a__f(a, b)
a__f(z0, z1) → f(z0, z1)
a__b → a
a__b → b
mark(f(z0, z1)) → a__f(mark(z0), z1)
mark(b) → a__b
mark(a) → a

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MARK(f(z0, z1)) → c4(MARK(z0))
S tuples:

MARK(f(z0, z1)) → c4(MARK(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

MARK

Compound Symbols:

c4

(9) 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(f(z0, z1)) → c4(MARK(z0))
We considered the (Usable) Rules:none
And the Tuples:

MARK(f(z0, z1)) → c4(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(MARK(x1)) = [5]x1   
POL(c4(x1)) = x1   
POL(f(x1, x2)) = [1] + x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MARK(f(z0, z1)) → c4(MARK(z0))
S tuples:none
K tuples:

MARK(f(z0, z1)) → c4(MARK(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

MARK

Compound Symbols:

c4

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

The set S is empty

(12) BOUNDS(1, 1)