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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
  ↳6 QDP
    ↳7 QDPOrderProof (⇔)
    ↳8 QDP
    ↳9 PisEmptyProof (⇔)
    ↳10 TRUE

(0) Obligation:

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

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

The TRS R consists of the following rules:

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(c) → a__c
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 2 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

A__F(b, X, c) → A__F(X, a__c, X)

The TRS R consists of the following rules:

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

(6) Obligation:

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

MARK(f(X1, X2, X3)) → MARK(X2)

The TRS R consists of the following rules:

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(c) → a__c
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) 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)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
f(x1, x2, x3)  =  f(x2)
a__f(x1, x2, x3)  =  a__f(x2)
b  =  b
c  =  c
a__c  =  a__c
mark(x1)  =  mark(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
mark1 > af1 > [f1, b, c, ac]

Status:
f1: [1]
af1: [1]
b: []
c: []
ac: []
mark1: [1]


The following usable rules [FROCOS05] were oriented:

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

(8) Obligation:

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

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(c) → a__c
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) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE