Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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.


↳ QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

MARK(b) → A__B
MARK(g(X)) → A__G(mark(X))
MARK(f(X1, X2, X3)) → A__F(X1, X2, X3)
A__F(X, g(X), Y) → A__F(Y, Y, Y)
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.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MARK(g(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(MARK(x1)) = (2)x_1   
POL(g(x1)) = 1/4 + (7/2)x_1   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

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