Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 QDP

(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: Combined order from the following AFS and order.
A__F(x1, x2, x3)  =  x3
MARK(x1)  =  x1
f(x1, x2, x3)  =  f(x3)
mark(x1)  =  x1
b  =  b
a  =  a
c  =  c
a__c  =  a__c
a__f(x1, x2, x3)  =  a__f(x3)

Recursive Path Order [RPO].
Precedence:
[f1, af1]
[c, ac] > b
[c, ac] > a


The following usable rules [FROCOS05] were oriented:

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

(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.