Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

a__f(X, X) → a__f(a, b)
a__b → a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
a__b → b

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__f(X, X) → a__f(a, b)
a__b → a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
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
A__F(X, X) → A__F(a, b)
MARK(f(X1, X2)) → MARK(X1)
MARK(f(X1, X2)) → A__F(mark(X1), X2)

The TRS R consists of the following rules:

a__f(X, X) → a__f(a, b)
a__b → a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
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
A__F(X, X) → A__F(a, b)
MARK(f(X1, X2)) → MARK(X1)
MARK(f(X1, X2)) → A__F(mark(X1), X2)

The TRS R consists of the following rules:

a__f(X, X) → a__f(a, b)
a__b → a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
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 1 SCC with 3 less nodes.

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

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

MARK(f(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__f(X, X) → a__f(a, b)
a__b → a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
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(f(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(MARK(x1)) = x_1   
POL(f(x1, x2)) = 1 + (2)x_1 + (2)x_2   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a__f(X, X) → a__f(a, b)
a__b → a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
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.