### (0) Obligation:

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

a__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(X) → h(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(z0)) → a__c(f(g(f(z0))))
a__f(z0) → f(z0)
a__c(z0) → d(z0)
a__c(z0) → c(z0)
a__h(z0) → a__c(d(z0))
a__h(z0) → h(z0)
mark(f(z0)) → a__f(mark(z0))
mark(c(z0)) → a__c(z0)
mark(h(z0)) → a__h(mark(z0))
mark(g(z0)) → g(z0)
mark(d(z0)) → d(z0)
Tuples:

A__F(f(z0)) → c1(A__C(f(g(f(z0)))))
A__F(z0) → c2
A__C(z0) → c3
A__C(z0) → c4
A__H(z0) → c5(A__C(d(z0)))
A__H(z0) → c6
MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0))
MARK(c(z0)) → c8(A__C(z0))
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0))
MARK(g(z0)) → c10
MARK(d(z0)) → c11
S tuples:

A__F(f(z0)) → c1(A__C(f(g(f(z0)))))
A__F(z0) → c2
A__C(z0) → c3
A__C(z0) → c4
A__H(z0) → c5(A__C(d(z0)))
A__H(z0) → c6
MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0))
MARK(c(z0)) → c8(A__C(z0))
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0))
MARK(g(z0)) → c10
MARK(d(z0)) → c11
K tuples:none
Defined Rule Symbols:

a__f, a__c, a__h, mark

Defined Pair Symbols:

A__F, A__C, A__H, MARK

Compound Symbols:

c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11

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

Removed 9 trailing nodes:

MARK(d(z0)) → c11
A__F(f(z0)) → c1(A__C(f(g(f(z0)))))
A__F(z0) → c2
A__H(z0) → c6
A__H(z0) → c5(A__C(d(z0)))
A__C(z0) → c3
A__C(z0) → c4
MARK(g(z0)) → c10
MARK(c(z0)) → c8(A__C(z0))

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(f(z0)) → a__c(f(g(f(z0))))
a__f(z0) → f(z0)
a__c(z0) → d(z0)
a__c(z0) → c(z0)
a__h(z0) → a__c(d(z0))
a__h(z0) → h(z0)
mark(f(z0)) → a__f(mark(z0))
mark(c(z0)) → a__c(z0)
mark(h(z0)) → a__h(mark(z0))
mark(g(z0)) → g(z0)
mark(d(z0)) → d(z0)
Tuples:

MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0))
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0))
S tuples:

MARK(f(z0)) → c7(A__F(mark(z0)), MARK(z0))
MARK(h(z0)) → c9(A__H(mark(z0)), MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, a__c, a__h, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c7, c9

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

Removed 2 trailing tuple parts

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(f(z0)) → a__c(f(g(f(z0))))
a__f(z0) → f(z0)
a__c(z0) → d(z0)
a__c(z0) → c(z0)
a__h(z0) → a__c(d(z0))
a__h(z0) → h(z0)
mark(f(z0)) → a__f(mark(z0))
mark(c(z0)) → a__c(z0)
mark(h(z0)) → a__h(mark(z0))
mark(g(z0)) → g(z0)
mark(d(z0)) → d(z0)
Tuples:

MARK(f(z0)) → c7(MARK(z0))
MARK(h(z0)) → c9(MARK(z0))
S tuples:

MARK(f(z0)) → c7(MARK(z0))
MARK(h(z0)) → c9(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, a__c, a__h, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c7, c9

### (7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

a__f(f(z0)) → a__c(f(g(f(z0))))
a__f(z0) → f(z0)
a__c(z0) → d(z0)
a__c(z0) → c(z0)
a__h(z0) → a__c(d(z0))
a__h(z0) → h(z0)
mark(f(z0)) → a__f(mark(z0))
mark(c(z0)) → a__c(z0)
mark(h(z0)) → a__h(mark(z0))
mark(g(z0)) → g(z0)
mark(d(z0)) → d(z0)

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MARK(f(z0)) → c7(MARK(z0))
MARK(h(z0)) → c9(MARK(z0))
S tuples:

MARK(f(z0)) → c7(MARK(z0))
MARK(h(z0)) → c9(MARK(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

MARK

Compound Symbols:

c7, c9

### (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)) → c7(MARK(z0))
MARK(h(z0)) → c9(MARK(z0))
We considered the (Usable) Rules:none
And the Tuples:

MARK(f(z0)) → c7(MARK(z0))
MARK(h(z0)) → c9(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(MARK(x1)) = [4]x1
POL(c7(x1)) = x1
POL(c9(x1)) = x1
POL(f(x1)) = [1] + x1
POL(h(x1)) = [4] + x1

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MARK(f(z0)) → c7(MARK(z0))
MARK(h(z0)) → c9(MARK(z0))
S tuples:none
K tuples:

MARK(f(z0)) → c7(MARK(z0))
MARK(h(z0)) → c9(MARK(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

MARK

Compound Symbols:

c7, c9

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

The set S is empty