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 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 DependencyGraphProof (⇔)
    ↳16 QDP
    ↳17 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 QDPOrderProof (⇔)
    ↳20 QDP
    ↳21 PisEmptyProof (⇔)
    ↳22 TRUE

(0) Obligation:

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

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(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:

A__FROM(X) → MARK(X)
A__LENGTH(cons(X, Y)) → A__LENGTH1(Y)
A__LENGTH1(X) → A__LENGTH(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(length(X)) → A__LENGTH(X)
MARK(length1(X)) → A__LENGTH1(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(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 2 SCCs with 2 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

A__LENGTH1(X) → A__LENGTH(X)
A__LENGTH(cons(X, Y)) → A__LENGTH1(Y)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(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.


A__LENGTH1(X) → A__LENGTH(X)
A__LENGTH(cons(X, Y)) → A__LENGTH1(Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__LENGTH1(x1)  =  A__LENGTH1(x1)
A__LENGTH(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [RPO].
Precedence:
[ALENGTH11, cons2]


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(X) → length1(X)

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(X) → length1(X)

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(from(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)
from(x1)  =  from(x1)
A__FROM(x1)  =  A__FROM(x1)
mark(x1)  =  mark(x1)
cons(x1, x2)  =  x1
s(x1)  =  x1
a__from(x1)  =  a__from(x1)
length(x1)  =  length
a__length(x1)  =  a__length
length1(x1)  =  length1
a__length1(x1)  =  a__length1
nil  =  nil
0  =  0

Recursive Path Order [RPO].
Precedence:
[from1, mark1, afrom1] > [MARK1, AFROM1] > [length, length1]
[from1, mark1, afrom1] > [alength, alength1, 0] > [length, length1]
[from1, mark1, afrom1] > nil > [length, length1]


The following usable rules [FROCOS05] were oriented:

mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length(X) → length(X)
a__length1(X) → a__length(X)
a__length1(X) → length1(X)

(12) Obligation:

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

MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(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.


MARK(from(X)) → A__FROM(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)
from(x1)  =  from(x1)
A__FROM(x1)  =  A__FROM(x1)
mark(x1)  =  x1
cons(x1, x2)  =  x1
s(x1)  =  x1
a__from(x1)  =  a__from(x1)
length(x1)  =  length
a__length(x1)  =  a__length
length1(x1)  =  length1
a__length1(x1)  =  a__length1
nil  =  nil
0  =  0

Recursive Path Order [RPO].
Precedence:
[from1, afrom1] > [MARK1, AFROM1, length, alength, length1, alength1, nil, 0]


The following usable rules [FROCOS05] were oriented:

mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length(X) → length(X)
a__length1(X) → a__length(X)
a__length1(X) → length1(X)

(14) Obligation:

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

A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(X) → length1(X)

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

(15) DependencyGraphProof (EQUIVALENT transformation)

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

(16) Obligation:

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

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

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(X) → length1(X)

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

(17) 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
s(x1)  =  s(x1)
cons(x1, x2)  =  x1

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(X) → length1(X)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(20) Obligation:

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

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(X) → length1(X)

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE