(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(0)) → 0
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))
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(0)) → 0
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) 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(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(0)) → 0
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.

(6) 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__F(a__p(s(0)))
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)  =  a__p
p(x1)  =  p

Recursive path order with status [RPO].
Precedence:
AF1 > ap > 0 > p
s1 > ap > 0 > p

Status:
p: multiset
ap: multiset
AF1: [1]
s1: multiset
0: multiset

The following usable rules [FROCOS05] were oriented:

a__p(s(0)) → 0
a__p(X) → p(X)

(7) 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(0)) → 0
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) 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(p(X)) → MARK(X)
MARK(f(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(0)) → 0
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.

(11) 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)
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
f(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)

Recursive path order with status [RPO].
Precedence:
trivial

Status:
cons2: multiset
s1: multiset

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

MARK(p(X)) → MARK(X)
MARK(f(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(0)) → 0
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.

(13) 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)  =  MARK(x1)
p(x1)  =  p(x1)
f(x1)  =  x1

Recursive path order with status [RPO].
Precedence:
p1 > MARK1

Status:
MARK1: [1]
p1: multiset

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

MARK(f(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(0)) → 0
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.

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(f(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Precedence:
f1 > MARK1

Status:
f1: multiset
MARK1: multiset

The following usable rules [FROCOS05] were oriented: none

(16) 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(0)) → 0
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.

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE