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 QDPOrderProof (⇔)
    ↳11 QDP
    ↳12 DependencyGraphProof (⇔)
    ↳13 QDP
    ↳14 QDPOrderProof (⇔)
    ↳15 QDP
    ↳16 PisEmptyProof (⇔)
    ↳17 TRUE
  ↳18 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))
MARK(f(X)) → A__F(mark(X))
MARK(f(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
A__P(x1)  =  x1
s(x1)  =  x1
0  =  0
MARK(x1)  =  x1
f(x1)  =  f(x1)
p(x1)  =  x1
mark(x1)  =  x1
cons(x1, x2)  =  x1
a__p(x1)  =  x1
a__f(x1)  =  a__f(x1)

Recursive Path Order [RPO].
Precedence:
[f1, af1] > AF > 0


The following usable rules [FROCOS05] were oriented:

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

(4) Obligation:

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

A__F(s(0)) → A__F(a__p(s(0)))
A__P(s(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.

(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)) → A__P(mark(X))
A__P(s(X)) → 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.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
p(x1)  =  x1
A__P(x1)  =  x1
mark(x1)  =  x1
s(x1)  =  x1
cons(x1, x2)  =  cons(x1)
a__p(x1)  =  x1
a__f(x1)  =  a__f
f(x1)  =  f
0  =  0

Recursive Path Order [RPO].
Precedence:
[af, f] > cons1
[af, f] > 0


The following usable rules [FROCOS05] were oriented:

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

(9) Obligation:

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

MARK(p(X)) → A__P(mark(X))
A__P(s(X)) → MARK(X)
MARK(p(X)) → MARK(X)
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.

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
[af, f] > cons


The following usable rules [FROCOS05] were oriented:

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

(11) Obligation:

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

MARK(p(X)) → A__P(mark(X))
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.

(12) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(13) 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.

(14) 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)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(15) 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.

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE

(18) 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.