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(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(b) → a__b
mark(a) → a
a__f(X1, X2, X3) → f(X1, X2, X3)
a__b → b

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(b) → a__b
mark(a) → a
a__f(X1, X2, X3) → f(X1, X2, X3)
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(a, X, X) → A__B
MARK(f(X1, X2, X3)) → MARK(X2)
MARK(f(X1, X2, X3)) → A__F(X1, mark(X2), X3)
A__F(a, X, X) → A__F(X, a__b, b)

The TRS R consists of the following rules:

a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(b) → a__b
mark(a) → a
a__f(X1, X2, X3) → f(X1, X2, X3)
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(a, X, X) → A__B
MARK(f(X1, X2, X3)) → MARK(X2)
MARK(f(X1, X2, X3)) → A__F(X1, mark(X2), X3)
A__F(a, X, X) → A__F(X, a__b, b)

The TRS R consists of the following rules:

a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(b) → a__b
mark(a) → a
a__f(X1, X2, X3) → f(X1, X2, X3)
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
            ↳ QDPOrderProof
          ↳ QDP

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

A__F(a, X, X) → A__F(X, a__b, b)

The TRS R consists of the following rules:

a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(b) → a__b
mark(a) → a
a__f(X1, X2, X3) → f(X1, X2, X3)
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.


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

POL(a) = 4   
POL(a__b) = 4   
POL(b) = 0   
POL(A__F(x1, x2, x3)) = (4)x_1 + (4)x_3   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none



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

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

a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(b) → a__b
mark(a) → a
a__f(X1, X2, X3) → f(X1, X2, X3)
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.

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

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

MARK(f(X1, X2, X3)) → MARK(X2)

The TRS R consists of the following rules:

a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(b) → a__b
mark(a) → a
a__f(X1, X2, X3) → f(X1, X2, X3)
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, X3)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(MARK(x1)) = (2)x_1   
POL(f(x1, x2, x3)) = 1/4 + (7/2)x_2   
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(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(b) → a__b
mark(a) → a
a__f(X1, X2, X3) → f(X1, X2, X3)
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.