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 (ComplexityIfPolyImplication, 0 ms)
↳4 CdtProblem
↳5 CdtUsableRulesProof (⇔, 0 ms)
↳6 CdtProblem
↳7 CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))), 42 ms)
↳8 CdtProblem
↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳10 BOUNDS(O(1), O(1))

(0) Obligation:

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

a__g(X) → a__h(X)
a__c → d
a__h(d) → a__g(c)
mark(g(X)) → a__g(X)
mark(h(X)) → a__h(X)
mark(c) → a__c
mark(d) → d
a__g(X) → g(X)
a__h(X) → h(X)
a__c → c

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

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

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

a__g, a__c, a__h, mark

Defined Pair Symbols:

A__G, A__C, A__H, MARK

Compound Symbols:

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

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 2 leading nodes:

MARK(h(z0)) → c8(A__H(z0))
MARK(g(z0)) → c7(A__G(z0))
Removed 6 trailing nodes:

A__G(z0) → c2
A__H(z0) → c6
A__C → c3
A__C → c4
MARK(d) → c10
MARK(c) → c9(A__C)

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
S tuples:

A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
K tuples:none
Defined Rule Symbols:

a__g, a__c, a__h, mark

Defined Pair Symbols:

A__G, A__H

Compound Symbols:

c1, c5

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

a__g(z0) → a__h(z0)
a__g(z0) → g(z0)
a__c → d
a__c → c
a__h(d) → a__g(c)
a__h(z0) → h(z0)
mark(g(z0)) → a__g(z0)
mark(h(z0)) → a__h(z0)
mark(c) → a__c
mark(d) → d

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
S tuples:

A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

A__G, A__H

Compound Symbols:

c1, c5

(7) 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.

A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
We considered the (Usable) Rules:none
And the Tuples:

A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(A__G(x1)) = [2] + [4]x1   
POL(A__H(x1)) = [4]x1   
POL(c) = 0   
POL(c1(x1)) = x1   
POL(c5(x1)) = x1   
POL(d) = [4]   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
S tuples:none
K tuples:

A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
Defined Rule Symbols:none

Defined Pair Symbols:

A__G, A__H

Compound Symbols:

c1, c5

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

The set S is empty

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