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

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

(0) Obligation:

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

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(X)) → mark(X)
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(X)

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(s(0)) → A__F(a__p(s(0)))
A__F(s(0)) → A__P(s(0))
A__P(s(X)) → MARK(X)
MARK(f(X)) → A__F(mark(X))
MARK(f(X)) → MARK(X)
MARK(p(X)) → A__P(mark(X))
MARK(p(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(X)) → mark(X)
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(X)

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(s(0)) → A__P(s(0))
A__P(s(X)) → MARK(X)
MARK(f(X)) → A__F(mark(X))
MARK(f(X)) → MARK(X)
MARK(p(X)) → A__P(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(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)  =  A__F(x1)
s(x1)  =  s(x1)
0  =  0
a__p(x1)  =  x1
A__P(x1)  =  A__P(x1)
MARK(x1)  =  MARK(x1)
f(x1)  =  f(x1)
mark(x1)  =  x1
p(x1)  =  x1
cons(x1, x2)  =  cons(x1)
a__f(x1)  =  a__f(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[s1, MARK1] > [AF1, f1, af1] > cons1 > [0, AP1]

Status:
cons1: [1]
f1: [1]
MARK1: [1]
af1: [1]
AF1: [1]
s1: [1]
AP1: [1]
0: []


The following usable rules [FROCOS05] were oriented:

a__p(X) → p(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
a__f(X) → f(X)
mark(s(X)) → s(mark(X))
a__f(s(0)) → a__f(a__p(s(0)))
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
a__p(s(X)) → mark(X)
a__f(0) → cons(0, f(s(0)))

(4) Obligation:

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

A__F(s(0)) → A__F(a__p(s(0)))
MARK(p(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(X)) → mark(X)
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(X)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.

(6) Complex Obligation (AND)

(7) Obligation:

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

MARK(p(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(X)) → mark(X)
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(X)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(p(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
p(x1)  =  p(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
p1: [1]


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(X)) → mark(X)
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(X)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

A__F(s(0)) → A__F(a__p(s(0)))

The TRS R consists of the following rules:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(X)) → mark(X)
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(X)

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