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 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP

(0) Obligation:

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

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

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

The TRS R consists of the following rules:

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

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(a, X, X) → A__F(X, a__b, b)

The TRS R consists of the following rules:

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


A__F(a, X, X) → A__F(X, a__b, b)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__F(x1, x2, x3)  =  A__F(x1, x3)
a  =  a
a__b  =  a__b
b  =  b
a__f(x1, x2, x3)  =  a__f(x3)
mark(x1)  =  mark(x1)
f(x1, x2, x3)  =  x3

Recursive Path Order [RPO].
Precedence:
mark1 > ab > a > AF2 > b
mark1 > af1 > b

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

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

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(b) → a__b
mark(a) → a
a__f(X1, X2, X3) → f(X1, X2, X3)
a__b → b

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