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

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

(0) Obligation:

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

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

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(g(X), Y) → A__F(mark(X), f(g(X), Y))
A__F(g(X), Y) → MARK(X)
MARK(f(X1, X2)) → A__F(mark(X1), X2)
MARK(f(X1, X2)) → MARK(X1)
MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive path order with status [RPO].
Quasi-Precedence:
g1 > [AF1, f1, MARK1, af1]

Status:
AF1: [1]
g1: multiset
f1: multiset
MARK1: [1]
af1: multiset


The following usable rules [FROCOS05] were oriented:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

(4) Obligation:

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

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

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

(5) PisEmptyProof (EQUIVALENT transformation)

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

(6) TRUE