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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 TRUE
  ↳22 QDP
    ↳23 QDPOrderProof (⇔)
    ↳24 QDP
    ↳25 PisEmptyProof (⇔)
    ↳26 TRUE
  ↳27 QDP
    ↳28 QDPOrderProof (⇔)
    ↳29 QDP
    ↳30 QDPOrderProof (⇔)
    ↳31 QDP
    ↳32 DependencyGraphProof (⇔)
    ↳33 AND
      ↳34 QDP
        ↳35 QDPOrderProof (⇔)
        ↳36 QDP
        ↳37 PisEmptyProof (⇔)
        ↳38 TRUE
      ↳39 QDP
        ↳40 QDPOrderProof (⇔)
        ↳41 QDP
        ↳42 QDPOrderProof (⇔)
        ↳43 QDP
        ↳44 PisEmptyProof (⇔)
        ↳45 TRUE

(0) Obligation:

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

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(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:

ACTIVE(f(0)) → MARK(cons(0, f(s(0))))
ACTIVE(f(0)) → CONS(0, f(s(0)))
ACTIVE(f(0)) → F(s(0))
ACTIVE(f(0)) → S(0)
ACTIVE(f(s(0))) → MARK(f(p(s(0))))
ACTIVE(f(s(0))) → F(p(s(0)))
ACTIVE(f(s(0))) → P(s(0))
ACTIVE(p(s(0))) → MARK(0)
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(f(X)) → F(mark(X))
MARK(f(X)) → MARK(X)
MARK(0) → ACTIVE(0)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(p(X)) → P(mark(X))
MARK(p(X)) → MARK(X)
F(mark(X)) → F(X)
F(active(X)) → F(X)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(X)
P(mark(X)) → P(X)
P(active(X)) → P(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(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 5 SCCs with 11 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

P(active(X)) → P(X)
P(mark(X)) → P(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(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.


P(active(X)) → P(X)
P(mark(X)) → P(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
P(x1)  =  x1
active(x1)  =  active(x1)
mark(x1)  =  mark(x1)
f(x1)  =  f
0  =  0
cons(x1, x2)  =  x2
s(x1)  =  s
p(x1)  =  p

Recursive Path Order [RPO].
Precedence:
s > p > [active1, mark1] > 0 > f


The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(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:

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(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.


S(active(X)) → S(X)
S(mark(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1)  =  x1
active(x1)  =  active(x1)
mark(x1)  =  mark(x1)
f(x1)  =  f
0  =  0
cons(x1, x2)  =  x2
s(x1)  =  s
p(x1)  =  p

Recursive Path Order [RPO].
Precedence:
s > p > [active1, mark1] > 0 > f


The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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

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

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

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

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
s > [CONS1, mark1, active1] > [f, 0, p]


The following usable rules [FROCOS05] were oriented:

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

(17) Obligation:

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

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

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

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
p > [f, 0] > cons > [mark1, active1] > CONS2
p > [f, 0] > s > [mark1, active1] > CONS2


The following usable rules [FROCOS05] were oriented:

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

(19) Obligation:

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

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

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

F(active(X)) → F(X)
F(mark(X)) → F(X)

The TRS R consists of the following rules:

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

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(active(X)) → F(X)
F(mark(X)) → F(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
active(x1)  =  active(x1)
mark(x1)  =  mark(x1)
f(x1)  =  f
0  =  0
cons(x1, x2)  =  x2
s(x1)  =  s
p(x1)  =  p

Recursive Path Order [RPO].
Precedence:
s > p > [active1, mark1] > 0 > f


The following usable rules [FROCOS05] were oriented:

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

(24) Obligation:

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

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

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(f(0)) → MARK(cons(0, f(s(0))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(f(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(p(X)) → MARK(X)

The TRS R consists of the following rules:

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

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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
cons(x1, x2)  =  x1
ACTIVE(x1)  =  x1
mark(x1)  =  x1
f(x1)  =  x1
0  =  0
s(x1)  =  s(x1)
p(x1)  =  x1
active(x1)  =  x1

Recursive Path Order [RPO].
Precedence:
[0, s1]


The following usable rules [FROCOS05] were oriented:

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

(29) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(f(0)) → MARK(cons(0, f(s(0))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(f(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(p(X)) → MARK(X)

The TRS R consists of the following rules:

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

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(f(0)) → MARK(cons(0, f(s(0))))
MARK(f(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)
cons(x1, x2)  =  x1
ACTIVE(x1)  =  ACTIVE(x1)
mark(x1)  =  x1
f(x1)  =  f(x1)
0  =  0
s(x1)  =  s(x1)
p(x1)  =  x1
active(x1)  =  x1

Recursive Path Order [RPO].
Precedence:
[MARK1, ACTIVE1, f1, 0, s1]


The following usable rules [FROCOS05] were oriented:

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

(31) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → MARK(X1)
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(p(X)) → MARK(X)

The TRS R consists of the following rules:

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

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

(32) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(33) Complex Obligation (AND)

(34) Obligation:

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

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(f(X)) → ACTIVE(f(mark(X)))

The TRS R consists of the following rules:

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

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

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
[0, p] > f1 > [s1, MARK1]
[0, p] > f1 > cons


The following usable rules [FROCOS05] were oriented:

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

(36) Obligation:

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

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

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

(37) PisEmptyProof (EQUIVALENT transformation)

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

(38) TRUE

(39) Obligation:

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

MARK(p(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

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

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

(40) 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)  =  MARK(x1)
p(x1)  =  x1
cons(x1, x2)  =  cons(x1)
active(x1)  =  x1
f(x1)  =  f(x1)
0  =  0
mark(x1)  =  x1
s(x1)  =  x1

Recursive Path Order [RPO].
Precedence:
0 > [cons1, f1] > MARK1


The following usable rules [FROCOS05] were oriented:

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

(41) 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:

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

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

(42) 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)
active(x1)  =  x1
f(x1)  =  f
0  =  0
mark(x1)  =  x1
cons(x1, x2)  =  x1
s(x1)  =  s

Recursive Path Order [RPO].
Precedence:
MARK1 > s
f > p1 > 0 > s


The following usable rules [FROCOS05] were oriented:

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

(43) Obligation:

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

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

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

(44) PisEmptyProof (EQUIVALENT transformation)

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

(45) TRUE