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 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 PisEmptyProof (⇔)
    ↳18 TRUE
  ↳19 QDP
    ↳20 QDPOrderProof (⇔)
    ↳21 QDP
    ↳22 QDPOrderProof (⇔)
    ↳23 QDP
    ↳24 QDPOrderProof (⇔)
    ↳25 QDP
    ↳26 PisEmptyProof (⇔)
    ↳27 TRUE
  ↳28 QDP
    ↳29 QDPOrderProof (⇔)
    ↳30 QDP
    ↳31 QDPOrderProof (⇔)
    ↳32 QDP
    ↳33 PisEmptyProof (⇔)
    ↳34 TRUE
  ↳35 QDP
    ↳36 QDPOrderProof (⇔)
    ↳37 QDP
    ↳38 DependencyGraphProof (⇔)
    ↳39 AND
      ↳40 QDP
        ↳41 QDPOrderProof (⇔)
        ↳42 QDP
        ↳43 PisEmptyProof (⇔)
        ↳44 TRUE
      ↳45 QDP
        ↳46 QDPOrderProof (⇔)
        ↳47 QDP
        ↳48 PisEmptyProof (⇔)
        ↳49 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(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)  =  x1
mark(x1)  =  mark(x1)
f(x1)  =  f(x1)
0  =  0
cons(x1, x2)  =  cons
s(x1)  =  s
p(x1)  =  p

Lexicographic path order with status [LPO].
Precedence:
s > p > 0 > mark1 > f1
s > p > 0 > cons

Status:
mark1: [1]
f1: [1]
0: []
cons: []
s: []
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)

(7) Obligation:

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

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.

(8) 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)
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)
f(x1)  =  f(x1)
0  =  0
mark(x1)  =  mark(x1)
cons(x1, x2)  =  cons
s(x1)  =  s
p(x1)  =  p

Lexicographic path order with status [LPO].
Precedence:
s > p > 0 > cons > mark1 > f1 > active1

Status:
active1: [1]
f1: [1]
0: []
mark1: [1]
cons: []
s: []
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)

(9) 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.

(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:

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.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Precedence:
s > p > 0 > mark1 > f1
s > p > 0 > cons

Status:
mark1: [1]
f1: [1]
0: []
cons: []
s: []
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)

(14) Obligation:

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

S(active(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.

(15) 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)
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)
f(x1)  =  f(x1)
0  =  0
mark(x1)  =  mark(x1)
cons(x1, x2)  =  cons
s(x1)  =  s
p(x1)  =  p

Lexicographic path order with status [LPO].
Precedence:
s > p > 0 > cons > mark1 > f1 > active1

Status:
active1: [1]
f1: [1]
0: []
mark1: [1]
cons: []
s: []
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)

(16) 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.

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) 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.

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

Lexicographic path order with status [LPO].
Precedence:
s > p > 0 > mark1
s > p > 0 > cons

Status:
CONS2: [2,1]
mark1: [1]
0: []
cons: []
s: []
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)

(21) Obligation:

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

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.

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Precedence:
s > p > 0 > mark1 > f1 > active1
s > p > 0 > cons > active1

Status:
active1: [1]
f1: [1]
0: []
mark1: [1]
cons: []
s: []
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)

(23) Obligation:

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

CONS(active(X1), 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.

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Precedence:
s > p > 0 > mark1 > active1 > f1
s > p > 0 > cons > active1 > f1

Status:
CONS2: [1,2]
active1: [1]
f1: [1]
0: []
mark1: [1]
cons: []
s: []
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)

(25) 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.

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE

(28) 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.

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Precedence:
s > p > 0 > mark1 > f1
s > p > 0 > cons

Status:
mark1: [1]
f1: [1]
0: []
cons: []
s: []
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)

(30) Obligation:

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

F(active(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.

(31) 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)
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)
f(x1)  =  f(x1)
0  =  0
mark(x1)  =  mark(x1)
cons(x1, x2)  =  cons
s(x1)  =  s
p(x1)  =  p

Lexicographic path order with status [LPO].
Precedence:
s > p > 0 > cons > mark1 > f1 > active1

Status:
active1: [1]
f1: [1]
0: []
mark1: [1]
cons: []
s: []
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)

(32) 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.

(33) PisEmptyProof (EQUIVALENT transformation)

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

(34) TRUE

(35) 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.

(36) 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)
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)  =  f(x1)
0  =  0
s(x1)  =  s(x1)
p(x1)  =  x1
active(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
0 > f1 > s1

Status:
f1: [1]
0: []
s1: [1]

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)

(37) 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.

(38) DependencyGraphProof (EQUIVALENT transformation)

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

(39) Complex Obligation (AND)

(40) 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.

(41) 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
0  =  0
MARK(x1)  =  MARK(x1)
p(x1)  =  p
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
cons(x1, x2)  =  cons

Lexicographic path order with status [LPO].
Precedence:
s > MARK1 > mark1 > f1 > active1
s > p > 0 > cons > mark1 > f1 > active1

Status:
f1: [1]
s: []
0: []
MARK1: [1]
p: []
mark1: [1]
active1: [1]
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)

(42) 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.

(43) PisEmptyProof (EQUIVALENT transformation)

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

(44) TRUE

(45) 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.

(46) 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)
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)  =  p(x1)
cons(x1, x2)  =  cons(x1)
active(x1)  =  x1
f(x1)  =  f
0  =  0
mark(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Precedence:
f > cons1 > p1
f > s1 > 0 > p1

Status:
p1: [1]
cons1: [1]
f: []
0: []
s1: [1]

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)

(47) 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.

(48) PisEmptyProof (EQUIVALENT transformation)

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

(49) TRUE