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 PisEmptyProof (⇔)
    ↳25 TRUE
  ↳26 QDP
    ↳27 QDPOrderProof (⇔)
    ↳28 QDP
    ↳29 QDPOrderProof (⇔)
    ↳30 QDP
    ↳31 QDPOrderProof (⇔)
    ↳32 QDP
    ↳33 QDPOrderProof (⇔)
    ↳34 QDP
    ↳35 PisEmptyProof (⇔)
    ↳36 TRUE
  ↳37 QDP
    ↳38 QDPOrderProof (⇔)
    ↳39 QDP
    ↳40 QDPOrderProof (⇔)
    ↳41 QDP
    ↳42 PisEmptyProof (⇔)
    ↳43 TRUE
  ↳44 QDP
    ↳45 QDPOrderProof (⇔)
    ↳46 QDP
    ↳47 QDPOrderProof (⇔)
    ↳48 QDP
    ↳49 QDPOrderProof (⇔)
    ↳50 QDP
    ↳51 QDPOrderProof (⇔)
    ↳52 QDP
    ↳53 PisEmptyProof (⇔)
    ↳54 TRUE

(0) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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(from(X)) → MARK(cons(X, from(s(X))))
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(length(nil)) → MARK(0)
ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
ACTIVE(length(cons(X, Y))) → S(length1(Y))
ACTIVE(length(cons(X, Y))) → LENGTH1(Y)
ACTIVE(length1(X)) → MARK(length(X))
ACTIVE(length1(X)) → LENGTH(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → FROM(mark(X))
MARK(from(X)) → MARK(X)
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(length(X)) → ACTIVE(length(X))
MARK(nil) → ACTIVE(nil)
MARK(0) → ACTIVE(0)
MARK(length1(X)) → ACTIVE(length1(X))
FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(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)
LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)
LENGTH1(mark(X)) → LENGTH1(X)
LENGTH1(active(X)) → LENGTH1(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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 6 SCCs with 12 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

LENGTH1(active(X)) → LENGTH1(X)
LENGTH1(mark(X)) → LENGTH1(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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.


LENGTH1(active(X)) → LENGTH1(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LENGTH1(x1)  =  LENGTH1(x1)
active(x1)  =  active(x1)
mark(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
[LENGTH11, active1]

Status:
active1: multiset
LENGTH11: multiset


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

LENGTH1(mark(X)) → LENGTH1(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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.


LENGTH1(mark(X)) → LENGTH1(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:
mark1 > LENGTH11

Status:
LENGTH11: multiset
mark1: multiset


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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:

LENGTH(active(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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.


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

Recursive path order with status [RPO].
Quasi-Precedence:
[LENGTH1, active1]

Status:
active1: multiset
LENGTH1: multiset


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

LENGTH(mark(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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.


LENGTH(mark(X)) → LENGTH(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:
mark1 > LENGTH1

Status:
LENGTH1: 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:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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:

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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.


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)  =  S(x1)
active(x1)  =  active(x1)
mark(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
[S1, active1]

Status:
active1: multiset
S1: multiset


The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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.


S(mark(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:
mark1 > S1

Status:
mark1: multiset
S1: multiset


The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) 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(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(27) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  CONS(x2)
mark(x1)  =  mark(x1)
active(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
[CONS1, mark1]

Status:
CONS1: multiset
mark1: multiset


The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

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(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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.


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

Recursive path order with status [RPO].
Quasi-Precedence:
mark1 > CONS2

Status:
CONS2: multiset
mark1: [1]


The following usable rules [FROCOS05] were oriented: none

(30) 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(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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.


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)

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

Status:
active1: multiset


The following usable rules [FROCOS05] were oriented: none

(32) 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(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(33) 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)
active(x1)  =  active(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
[CONS1, active1]

Status:
active1: [1]
CONS1: multiset


The following usable rules [FROCOS05] were oriented: none

(34) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) TRUE

(37) Obligation:

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

FROM(active(X)) → FROM(X)
FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(38) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive path order with status [RPO].
Quasi-Precedence:
[FROM1, active1]

Status:
active1: multiset
FROM1: multiset


The following usable rules [FROCOS05] were oriented: none

(39) Obligation:

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

FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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.


FROM(mark(X)) → FROM(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:
mark1 > FROM1

Status:
mark1: multiset
FROM1: multiset


The following usable rules [FROCOS05] were oriented: none

(41) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) TRUE

(44) Obligation:

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

MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(length1(X)) → MARK(length(X))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(X))
MARK(length1(X)) → ACTIVE(length1(X))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → ACTIVE(s(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK
from(x1)  =  from
ACTIVE(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  cons
s(x1)  =  s
length(x1)  =  length
length1(x1)  =  length1
active(x1)  =  x1
nil  =  nil
0  =  0

Recursive path order with status [RPO].
Quasi-Precedence:
[MARK, from, length, length1] > cons > [mark, s]
[MARK, from, length, length1] > 0 > [mark, s]
nil > 0 > [mark, s]

Status:
from: []
MARK: []
cons: multiset
length1: []
mark: multiset
s: multiset
length: []
0: multiset
nil: multiset


The following usable rules [FROCOS05] were oriented:

from(active(X)) → from(X)
from(mark(X)) → from(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
length1(active(X)) → length1(X)
length1(mark(X)) → length1(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)

(46) Obligation:

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

MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(X)) → MARK(X)
ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(length1(X)) → MARK(length(X))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(X))
MARK(length1(X)) → ACTIVE(length1(X))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(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
from(x1)  =  from(x1)
ACTIVE(x1)  =  x1
mark(x1)  =  x1
cons(x1, x2)  =  cons(x1)
s(x1)  =  x1
length(x1)  =  length
length1(x1)  =  length1
active(x1)  =  x1
nil  =  nil
0  =  0

Recursive path order with status [RPO].
Quasi-Precedence:
from1 > cons1
[length, length1] > 0

Status:
from1: [1]
cons1: multiset
length1: []
length: []
0: multiset
nil: multiset


The following usable rules [FROCOS05] were oriented:

from(active(X)) → from(X)
from(mark(X)) → from(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
mark(nil) → active(nil)
mark(0) → active(0)
active(from(X)) → mark(cons(X, from(s(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(length1(X)) → mark(length(X))
mark(s(X)) → active(s(mark(X)))
active(length(cons(X, Y))) → mark(s(length1(Y)))
mark(from(X)) → active(from(mark(X)))
mark(length1(X)) → active(length1(X))
mark(length(X)) → active(length(X))
active(length(nil)) → mark(0)
length(active(X)) → length(X)
length(mark(X)) → length(X)
length1(active(X)) → length1(X)
length1(mark(X)) → length1(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)

(48) Obligation:

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

MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
ACTIVE(length1(X)) → MARK(length(X))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(X))
MARK(length1(X)) → ACTIVE(length1(X))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
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
from(x1)  =  from
ACTIVE(x1)  =  x1
mark(x1)  =  mark(x1)
length(x1)  =  x1
cons(x1, x2)  =  cons(x2)
s(x1)  =  s(x1)
length1(x1)  =  x1
active(x1)  =  x1
nil  =  nil
0  =  0

Recursive path order with status [RPO].
Quasi-Precedence:
from > cons1 > [mark1, s1]
nil > [mark1, s1]
0 > [mark1, s1]

Status:
from: []
cons1: [1]
mark1: multiset
s1: [1]
0: multiset
nil: multiset


The following usable rules [FROCOS05] were oriented:

from(active(X)) → from(X)
from(mark(X)) → from(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
length1(active(X)) → length1(X)
length1(mark(X)) → length1(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)

(50) Obligation:

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

MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(length1(X)) → MARK(length(X))
MARK(length(X)) → ACTIVE(length(X))
MARK(length1(X)) → ACTIVE(length1(X))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(51) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(length1(X)) → MARK(length(X))
MARK(length(X)) → ACTIVE(length(X))
MARK(length1(X)) → ACTIVE(length1(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
from(x1)  =  x1
ACTIVE(x1)  =  x1
mark(x1)  =  x1
length1(x1)  =  length1(x1)
length(x1)  =  x1
active(x1)  =  x1
cons(x1, x2)  =  x2
nil  =  nil
0  =  0
s(x1)  =  s

Recursive path order with status [RPO].
Quasi-Precedence:
length11 > MARK1 > s
[nil, 0] > s

Status:
MARK1: [1]
length11: [1]
s: []
0: multiset
nil: multiset


The following usable rules [FROCOS05] were oriented:

from(active(X)) → from(X)
from(mark(X)) → from(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
mark(nil) → active(nil)
mark(0) → active(0)
length(active(X)) → length(X)
length(mark(X)) → length(X)
active(from(X)) → mark(cons(X, from(s(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(length1(X)) → mark(length(X))
mark(s(X)) → active(s(mark(X)))
active(length(cons(X, Y))) → mark(s(length1(Y)))
mark(from(X)) → active(from(mark(X)))
mark(length1(X)) → active(length1(X))
mark(length(X)) → active(length(X))
length1(active(X)) → length1(X)
length1(mark(X)) → length1(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
active(length(nil)) → mark(0)

(52) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(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)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(53) PisEmptyProof (EQUIVALENT transformation)

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

(54) TRUE