Termination w.r.t. Q of the given QTRS could be disproven:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(4) Obligation:

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

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

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.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( A__F(x₁, ..., x₃) ) = x₃ + 2

POL( mark(x₁) ) = x₁

POL( f(x₁, ..., x₃) ) = x₃ + 1

POL( a__f(x₁, ..., x₃) ) = x₃ + 1

POL( c ) = 1

POL( a__c ) = 1

POL( a ) = 0

POL( b ) = 0

POL( MARK(x₁) ) = x₁ + 2

The following usable rules [FROCOS05] were oriented:

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__f(a, b, X) → a__f(X, X, mark(X))
a__c → a
a__c → b
a__c → c

(6) Obligation:

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

A__F(a, b, X) → MARK(X)
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.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(8) Obligation:

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.

(9) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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:

• Matcher: [ ]
• Semiunifier: [ ]

---

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.

(10) NO