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__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)
mark1(g1(X)) -> g1(mark1(X))
a__f2(X1, X2) -> f2(X1, X2)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)
mark1(g1(X)) -> g1(mark1(X))
a__f2(X1, X2) -> f2(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:

MARK1(f2(X1, X2)) -> A__F2(mark1(X1), X2)
A__F2(g1(X), Y) -> MARK1(X)
A__F2(g1(X), Y) -> A__F2(mark1(X), f2(g1(X), Y))
MARK1(g1(X)) -> MARK1(X)
MARK1(f2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)
mark1(g1(X)) -> g1(mark1(X))
a__f2(X1, X2) -> f2(X1, X2)

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPOrderProof

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

MARK1(f2(X1, X2)) -> A__F2(mark1(X1), X2)
A__F2(g1(X), Y) -> MARK1(X)
A__F2(g1(X), Y) -> A__F2(mark1(X), f2(g1(X), Y))
MARK1(g1(X)) -> MARK1(X)
MARK1(f2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)
mark1(g1(X)) -> g1(mark1(X))
a__f2(X1, X2) -> f2(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.


MARK1(f2(X1, X2)) -> A__F2(mark1(X1), X2)
MARK1(f2(X1, X2)) -> MARK1(X1)
The remaining pairs can at least be oriented weakly.

A__F2(g1(X), Y) -> MARK1(X)
A__F2(g1(X), Y) -> A__F2(mark1(X), f2(g1(X), Y))
MARK1(g1(X)) -> MARK1(X)
Used ordering: Polynomial interpretation [21]:

POL(A__F2(x1, x2)) = x1   
POL(MARK1(x1)) = x1   
POL(a__f2(x1, x2)) = 1 + x1   
POL(f2(x1, x2)) = 1 + x1   
POL(g1(x1)) = x1   
POL(mark1(x1)) = x1   

The following usable rules [14] were oriented:

mark1(g1(X)) -> g1(mark1(X))
a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
a__f2(X1, X2) -> f2(X1, X2)
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ DependencyGraphProof

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

A__F2(g1(X), Y) -> MARK1(X)
A__F2(g1(X), Y) -> A__F2(mark1(X), f2(g1(X), Y))
MARK1(g1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)
mark1(g1(X)) -> g1(mark1(X))
a__f2(X1, X2) -> f2(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 2 SCCs with 1 less node.

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

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

MARK1(g1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)
mark1(g1(X)) -> g1(mark1(X))
a__f2(X1, X2) -> f2(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.


MARK1(g1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MARK1(x1)) = x1   
POL(g1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)
mark1(g1(X)) -> g1(mark1(X))
a__f2(X1, X2) -> f2(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.

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

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

A__F2(g1(X), Y) -> A__F2(mark1(X), f2(g1(X), Y))

The TRS R consists of the following rules:

a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)
mark1(g1(X)) -> g1(mark1(X))
a__f2(X1, X2) -> f2(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__F2(g1(X), Y) -> A__F2(mark1(X), f2(g1(X), Y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(A__F2(x1, x2)) = x1   
POL(a__f2(x1, x2)) = x2   
POL(f2(x1, x2)) = x2   
POL(g1(x1)) = 1 + x1   
POL(mark1(x1)) = x1   

The following usable rules [14] were oriented:

mark1(g1(X)) -> g1(mark1(X))
a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
a__f2(X1, X2) -> f2(X1, X2)
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)



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

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

a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)
mark1(g1(X)) -> g1(mark1(X))
a__f2(X1, X2) -> f2(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.