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__h(X) → a__g(mark(X), X)
a__g(a, X) → a__f(b, X)
a__f(X, X) → a__h(a__a)
a__a → b
mark(h(X)) → a__h(mark(X))
mark(g(X1, X2)) → a__g(mark(X1), X2)
mark(a) → a__a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → b
a__h(X) → h(X)
a__g(X1, X2) → g(X1, X2)
a__a → a
a__f(X1, X2) → f(X1, X2)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

a__h(X) → a__g(mark(X), X)
a__g(a, X) → a__f(b, X)
a__f(X, X) → a__h(a__a)
a__a → b
mark(h(X)) → a__h(mark(X))
mark(g(X1, X2)) → a__g(mark(X1), X2)
mark(a) → a__a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → b
a__h(X) → h(X)
a__g(X1, X2) → g(X1, X2)
a__a → a
a__f(X1, X2) → f(X1, X2)

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(h(X)) → A__H(mark(X))
A__F(X, X) → A__A
MARK(f(X1, X2)) → MARK(X1)
A__H(X) → A__G(mark(X), X)
MARK(a) → A__A
MARK(h(X)) → MARK(X)
A__G(a, X) → A__F(b, X)
A__H(X) → MARK(X)
MARK(g(X1, X2)) → A__G(mark(X1), X2)
MARK(f(X1, X2)) → A__F(mark(X1), X2)
MARK(g(X1, X2)) → MARK(X1)
A__F(X, X) → A__H(a__a)

The TRS R consists of the following rules:

a__h(X) → a__g(mark(X), X)
a__g(a, X) → a__f(b, X)
a__f(X, X) → a__h(a__a)
a__a → b
mark(h(X)) → a__h(mark(X))
mark(g(X1, X2)) → a__g(mark(X1), X2)
mark(a) → a__a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → b
a__h(X) → h(X)
a__g(X1, X2) → g(X1, X2)
a__a → a
a__f(X1, X2) → f(X1, X2)

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(h(X)) → A__H(mark(X))
A__F(X, X) → A__A
MARK(f(X1, X2)) → MARK(X1)
A__H(X) → A__G(mark(X), X)
MARK(a) → A__A
MARK(h(X)) → MARK(X)
A__G(a, X) → A__F(b, X)
A__H(X) → MARK(X)
MARK(g(X1, X2)) → A__G(mark(X1), X2)
MARK(f(X1, X2)) → A__F(mark(X1), X2)
MARK(g(X1, X2)) → MARK(X1)
A__F(X, X) → A__H(a__a)

The TRS R consists of the following rules:

a__h(X) → a__g(mark(X), X)
a__g(a, X) → a__f(b, X)
a__f(X, X) → a__h(a__a)
a__a → b
mark(h(X)) → a__h(mark(X))
mark(g(X1, X2)) → a__g(mark(X1), X2)
mark(a) → a__a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → b
a__h(X) → h(X)
a__g(X1, X2) → g(X1, X2)
a__a → a
a__f(X1, X2) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof

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

MARK(h(X)) → A__H(mark(X))
MARK(f(X1, X2)) → MARK(X1)
A__H(X) → A__G(mark(X), X)
MARK(h(X)) → MARK(X)
A__G(a, X) → A__F(b, X)
A__H(X) → MARK(X)
MARK(g(X1, X2)) → A__G(mark(X1), X2)
MARK(f(X1, X2)) → A__F(mark(X1), X2)
MARK(g(X1, X2)) → MARK(X1)
A__F(X, X) → A__H(a__a)

The TRS R consists of the following rules:

a__h(X) → a__g(mark(X), X)
a__g(a, X) → a__f(b, X)
a__f(X, X) → a__h(a__a)
a__a → b
mark(h(X)) → a__h(mark(X))
mark(g(X1, X2)) → a__g(mark(X1), X2)
mark(a) → a__a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → b
a__h(X) → h(X)
a__g(X1, X2) → g(X1, X2)
a__a → a
a__f(X1, X2) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MARK(h(X)) → A__H(mark(X))
MARK(f(X1, X2)) → MARK(X1)
MARK(h(X)) → MARK(X)
MARK(g(X1, X2)) → A__G(mark(X1), X2)
MARK(f(X1, X2)) → A__F(mark(X1), X2)
MARK(g(X1, X2)) → MARK(X1)


Used ordering: POLO with Polynomial interpretation [25]:

POL(A__F(x1, x2)) = 2·x1 + x2   
POL(A__G(x1, x2)) = x1 + x2   
POL(A__H(x1)) = 2·x1   
POL(MARK(x1)) = x1   
POL(a) = 0   
POL(a__a) = 0   
POL(a__f(x1, x2)) = 1 + 2·x1 + x2   
POL(a__g(x1, x2)) = 1 + x1 + x2   
POL(a__h(x1)) = 1 + 2·x1   
POL(b) = 0   
POL(f(x1, x2)) = 1 + 2·x1 + x2   
POL(g(x1, x2)) = 1 + x1 + x2   
POL(h(x1)) = 1 + 2·x1   
POL(mark(x1)) = x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ DependencyGraphProof

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

A__H(X) → A__G(mark(X), X)
A__G(a, X) → A__F(b, X)
A__H(X) → MARK(X)
A__F(X, X) → A__H(a__a)

The TRS R consists of the following rules:

a__h(X) → a__g(mark(X), X)
a__g(a, X) → a__f(b, X)
a__f(X, X) → a__h(a__a)
a__a → b
mark(h(X)) → a__h(mark(X))
mark(g(X1, X2)) → a__g(mark(X1), X2)
mark(a) → a__a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → b
a__h(X) → h(X)
a__g(X1, X2) → g(X1, X2)
a__a → a
a__f(X1, X2) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ Instantiation

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

A__H(X) → A__G(mark(X), X)
A__G(a, X) → A__F(b, X)
A__F(X, X) → A__H(a__a)

The TRS R consists of the following rules:

a__h(X) → a__g(mark(X), X)
a__g(a, X) → a__f(b, X)
a__f(X, X) → a__h(a__a)
a__a → b
mark(h(X)) → a__h(mark(X))
mark(g(X1, X2)) → a__g(mark(X1), X2)
mark(a) → a__a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → b
a__h(X) → h(X)
a__g(X1, X2) → g(X1, X2)
a__a → a
a__f(X1, X2) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule A__F(X, X) → A__H(a__a) we obtained the following new rules:

A__F(b, b) → A__H(a__a)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP

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

A__H(X) → A__G(mark(X), X)
A__F(b, b) → A__H(a__a)
A__G(a, X) → A__F(b, X)

The TRS R consists of the following rules:

a__h(X) → a__g(mark(X), X)
a__g(a, X) → a__f(b, X)
a__f(X, X) → a__h(a__a)
a__a → b
mark(h(X)) → a__h(mark(X))
mark(g(X1, X2)) → a__g(mark(X1), X2)
mark(a) → a__a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → b
a__h(X) → h(X)
a__g(X1, X2) → g(X1, X2)
a__a → a
a__f(X1, X2) → f(X1, X2)

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