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))), 34 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(X, g(X), Y) → a__f(Y, Y, Y)
a__g(b) → c
a__b → c
mark(f(X1, X2, X3)) → a__f(X1, X2, X3)
mark(g(X)) → a__g(mark(X))
mark(b) → a__b
mark(c) → c
a__f(X1, X2, X3) → f(X1, X2, X3)
a__g(X) → g(X)
a__b → b

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0, g(z0), z1) → a__f(z1, z1, z1)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__g(b) → c
a__g(z0) → g(z0)
a__b → c
a__b → b
mark(f(z0, z1, z2)) → a__f(z0, z1, z2)
mark(g(z0)) → a__g(mark(z0))
mark(b) → a__b
mark(c) → c
Tuples:

A__F(z0, g(z0), z1) → c1(A__F(z1, z1, z1))
MARK(f(z0, z1, z2)) → c7(A__F(z0, z1, z2))
MARK(g(z0)) → c8(A__G(mark(z0)), MARK(z0))
MARK(b) → c9(A__B)
S tuples:

A__F(z0, g(z0), z1) → c1(A__F(z1, z1, z1))
MARK(f(z0, z1, z2)) → c7(A__F(z0, z1, z2))
MARK(g(z0)) → c8(A__G(mark(z0)), MARK(z0))
MARK(b) → c9(A__B)
K tuples:none
Defined Rule Symbols:

a__f, a__g, a__b, mark

Defined Pair Symbols:

A__F, MARK

Compound Symbols:

c1, c7, c8, c9

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

Removed 3 trailing nodes:

MARK(b) → c9(A__B)
A__F(z0, g(z0), z1) → c1(A__F(z1, z1, z1))
MARK(f(z0, z1, z2)) → c7(A__F(z0, z1, z2))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0, g(z0), z1) → a__f(z1, z1, z1)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__g(b) → c
a__g(z0) → g(z0)
a__b → c
a__b → b
mark(f(z0, z1, z2)) → a__f(z0, z1, z2)
mark(g(z0)) → a__g(mark(z0))
mark(b) → a__b
mark(c) → c
Tuples:

MARK(g(z0)) → c8(A__G(mark(z0)), MARK(z0))
S tuples:

MARK(g(z0)) → c8(A__G(mark(z0)), MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, a__g, a__b, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c8

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

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0, g(z0), z1) → a__f(z1, z1, z1)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__g(b) → c
a__g(z0) → g(z0)
a__b → c
a__b → b
mark(f(z0, z1, z2)) → a__f(z0, z1, z2)
mark(g(z0)) → a__g(mark(z0))
mark(b) → a__b
mark(c) → c
Tuples:

MARK(g(z0)) → c8(MARK(z0))
S tuples:

MARK(g(z0)) → c8(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, a__g, a__b, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c8

(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(g(z0)) → c8(MARK(z0))
We considered the (Usable) Rules:none
And the Tuples:

MARK(g(z0)) → c8(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(MARK(x1)) = [5]x1   
POL(c8(x1)) = x1   
POL(g(x1)) = [1] + x1   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0, g(z0), z1) → a__f(z1, z1, z1)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__g(b) → c
a__g(z0) → g(z0)
a__b → c
a__b → b
mark(f(z0, z1, z2)) → a__f(z0, z1, z2)
mark(g(z0)) → a__g(mark(z0))
mark(b) → a__b
mark(c) → c
Tuples:

MARK(g(z0)) → c8(MARK(z0))
S tuples:none
K tuples:

MARK(g(z0)) → c8(MARK(z0))
Defined Rule Symbols:

a__f, a__g, a__b, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c8

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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