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(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
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__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MARK(f(X1, X2)) → A__F(mark(X1), X2)
A__F(g(X), Y) → MARK(X)
A__F(g(X), Y) → A__F(mark(X), f(g(X), Y))
MARK(g(X)) → MARK(X)
MARK(f(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

MARK(f(X1, X2)) → A__F(mark(X1), X2)
A__F(g(X), Y) → MARK(X)
A__F(g(X), Y) → A__F(mark(X), f(g(X), Y))
MARK(g(X)) → MARK(X)
MARK(f(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof

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

MARK(f(X1, X2)) → A__F(mark(X1), X2)
A__F(g(X), Y) → MARK(X)
A__F(g(X), Y) → A__F(mark(X), f(g(X), Y))
MARK(g(X)) → MARK(X)
MARK(f(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


A__F(g(X), Y) → MARK(X)
A__F(g(X), Y) → A__F(mark(X), f(g(X), Y))
MARK(g(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.

MARK(f(X1, X2)) → A__F(mark(X1), X2)
MARK(f(X1, X2)) → MARK(X1)
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
f(x1, x2)  =  x1
A__F(x1, x2)  =  x1
mark(x1)  =  x1
g(x1)  =  g(x1)
a__f(x1, x2)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
a__f(X1, X2) → f(X1, X2)
mark(g(X)) → g(mark(X))
mark(f(X1, X2)) → a__f(mark(X1), X2)



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

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

MARK(f(X1, X2)) → A__F(mark(X1), X2)
MARK(f(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
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 [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ 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(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


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: Combined order from the following AFS and order.
MARK(x1)  =  x1
f(x1, x2)  =  f(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

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.