Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

Q restricted rewrite system:
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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A__F(X, g(X), Y) → A__F(Y, Y, Y)
MARK(f(X1, X2, X3)) → A__F(X1, X2, X3)
MARK(g(X)) → A__G(mark(X))
MARK(g(X)) → MARK(X)
MARK(b) → A__B

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A__F(X, g(X), Y) → A__F(Y, Y, Y)

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MARK(g(X)) → MARK(X)

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(g(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
g(x1)  =  g(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

Q DP problem:
P is empty.
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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE