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__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 CpxTRS 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__H(d) → c5(A__G(c))
MARK(g(z0)) → c7(A__G(z0))
MARK(h(z0)) → c8(A__H(z0))
MARK(c) → c9(A__C)

S tuples:

A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
MARK(g(z0)) → c7(A__G(z0))
MARK(h(z0)) → c8(A__H(z0))
MARK(c) → c9(A__C)

K tuples:none
Defined Rule Symbols:

a__g, a__c, a__h, mark

Defined Pair Symbols:

A__G, A__H, MARK

Compound Symbols:

c1, c5, c7, c8, c9

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 2 leading nodes:

MARK(h(z0)) → c8(A__H(z0))
MARK(g(z0)) → c7(A__G(z0))

Removed 1 trailing nodes:

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) 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.

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]   

(6) 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:none
K tuples:

A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))

Defined Rule Symbols:

a__g, a__c, a__h, mark

Defined Pair Symbols:

A__G, A__H

Compound Symbols:

c1, c5

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))