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

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

(0) Obligation:

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

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

The TRS R consists of the following rules:

a__f(X, X) → a__f(a, b)
a__b → a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
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 1 SCC with 3 less nodes.

(4) Obligation:

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

MARK(f(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

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

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)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:
MARK1 > [f2, af2, a, b, ab, mark1]

Status:
MARK1: multiset
f2: [2,1]
af2: [2,1]
a: multiset
b: multiset
ab: multiset
mark1: [1]


The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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

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

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

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE