Termination w.r.t. Q proof of /tmp/tmpZRDSnd/Ex1_GM99

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

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

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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(f(X1, X2, X3)) → MARK(X3)
A__F(a, b, X) → A__F(X, X, mark(X))
MARK(f(X1, X2, X3)) → A__F(X1, X2, mark(X3))
A__F(a, b, X) → MARK(X)
MARK(c) → A__C

The TRS R consists of the following rules:

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

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(f(X1, X2, X3)) → MARK(X3)
A__F(a, b, X) → A__F(X, X, mark(X))
MARK(f(X1, X2, X3)) → A__F(X1, X2, mark(X3))
A__F(a, b, X) → MARK(X)
MARK(c) → A__C

The TRS R consists of the following rules:

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

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

          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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(X3)
MARK(f(X1, X2, X3)) → A__F(X1, X2, mark(X3))

The remaining pairs can at least be oriented weakly.

A__F(a, b, X) → A__F(X, X, mark(X))
A__F(a, b, X) → MARK(X)

Used ordering: Polynomial interpretation [25]:

POL(A__F(x₁, x₂, x₃)) = x₃
POL(MARK(x₁)) = x₁
POL(a) = 0
POL(a__c) = 0
POL(a__f(x₁, x₂, x₃)) = 1 + x₃
POL(b) = 0
POL(c) = 0
POL(f(x₁, x₂, x₃)) = 1 + x₃
POL(mark(x₁)) = x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

A__F(a, b, X) → A__F(X, X, mark(X))
A__F(a, b, X) → MARK(X)

The TRS R consists of the following rules:

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

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

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ NonTerminationProof

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

A__F(a, b, X) → A__F(X, X, mark(X))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

A__F(a, b, X) → A__F(X, X, mark(X))

The TRS R consists of the following rules:

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

s = A__F(mark(a__c), mark(a__c), mark(mark(a__c))) evaluates to t =A__F(mark(a__c), mark(a__c), mark(mark(a__c)))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

A__F(mark(a__c), mark(a__c), mark(mark(a__c))) → A__F(mark(a__c), mark(a__c), mark(mark(c)))
with rule a__c → c at position [2,0,0] and matcher [ ]

A__F(mark(a__c), mark(a__c), mark(mark(c))) → A__F(mark(a__c), mark(a__c), mark(a__c))
with rule mark(c) → a__c at position [2,0] and matcher [ ]

A__F(mark(a__c), mark(a__c), mark(a__c)) → A__F(mark(a__c), mark(a__c), mark(c))
with rule a__c → c at position [2,0] and matcher [ ]

A__F(mark(a__c), mark(a__c), mark(c)) → A__F(mark(a__c), mark(a__c), a__c)
with rule mark(c) → a__c at position [2] and matcher [ ]

A__F(mark(a__c), mark(a__c), a__c) → A__F(mark(a__c), mark(b), a__c)
with rule a__c → b at position [1,0] and matcher [ ]

A__F(mark(a__c), mark(b), a__c) → A__F(mark(a__c), b, a__c)
with rule mark(b) → b at position [1] and matcher [ ]

A__F(mark(a__c), b, a__c) → A__F(mark(a), b, a__c)
with rule a__c → a at position [0,0] and matcher [ ]

A__F(mark(a), b, a__c) → A__F(a, b, a__c)
with rule mark(a) → a at position [0] and matcher [ ]

A__F(a, b, a__c) → A__F(a__c, a__c, mark(a__c))
with rule A__F(a, b, X') → A__F(X', X', mark(X')) at position [] and matcher [X' / a__c]

A__F(a__c, a__c, mark(a__c)) → A__F(a__c, b, mark(a__c))
with rule a__c → b at position [1] and matcher [ ]

A__F(a__c, b, mark(a__c)) → A__F(a, b, mark(a__c))
with rule a__c → a at position [0] and matcher [ ]

A__F(a, b, mark(a__c)) → A__F(mark(a__c), mark(a__c), mark(mark(a__c)))
with rule A__F(a, b, X) → A__F(X, X, mark(X))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.