(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(mark(X)) → LENGTH1(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
LENGTH1(x0, x1)  =  LENGTH1(x1)

Tags:
LENGTH1 has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(LENGTH1(x1)) = 1   
POL(active(x1)) = x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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.

(8) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
LENGTH1(x0, x1)  =  LENGTH1(x1)

Tags:
LENGTH1 has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(LENGTH1(x1)) = 1   
POL(active(x1)) = 1 + x1   

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(mark(X)) → LENGTH(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
LENGTH(x0, x1)  =  LENGTH(x1)

Tags:
LENGTH has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(LENGTH(x1)) = 1   
POL(active(x1)) = x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

LENGTH(active(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(active(X)) → LENGTH(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
LENGTH(x0, x1)  =  LENGTH(x1)

Tags:
LENGTH has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(LENGTH(x1)) = 1   
POL(active(x1)) = 1 + x1   

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(mark(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
S(x0, x1)  =  S(x1)

Tags:
S has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(S(x1)) = 1   
POL(active(x1)) = x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(21) 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(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(active(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
S(x0, x1)  =  S(x1)

Tags:
S has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(S(x1)) = 1   
POL(active(x1)) = 1 + x1   

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(mark(X1), X2) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
CONS(x0, x1, x2)  =  CONS(x1)

Tags:
CONS has argument tags [1,0,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(CONS(x1, x2)) = 1   
POL(active(x1)) = x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

CONS(X1, mark(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(active(X1), X2) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
CONS(x0, x1, x2)  =  CONS(x1)

Tags:
CONS has argument tags [3,0,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(CONS(x1, x2)) = 1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = 0   

The following usable rules [FROCOS05] were oriented: none

(30) 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(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: SCNP Order with the following components:
Level mapping:
Top level AFS:
CONS(x0, x1, x2)  =  CONS(x0, x2)

Tags:
CONS has argument tags [0,3,3] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(CONS(x1, x2)) = 1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = x1   

The following usable rules [FROCOS05] were oriented: none

(32) Obligation:

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

CONS(X1, mark(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(X1, mark(X2)) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
CONS(x0, x1, x2)  =  CONS(x0, x1)

Tags:
CONS has argument tags [1,0,0] and root tag 0

Comparison: DMS
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(CONS(x1, x2)) = 1 + x2   
POL(mark(x1)) = 1 + x1   

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(mark(X)) → FROM(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
FROM(x0, x1)  =  FROM(x1)

Tags:
FROM has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(FROM(x1)) = 1   
POL(active(x1)) = x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(39) Obligation:

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

FROM(active(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(active(X)) → FROM(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
FROM(x0, x1)  =  FROM(x1)

Tags:
FROM has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(FROM(x1)) = 1   
POL(active(x1)) = 1 + x1   

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.


ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x0, x1)
ACTIVE(x0, x1)  =  ACTIVE(x0, x1)

Tags:
MARK has argument tags [1,1] and root tag 0
ACTIVE has argument tags [0,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = 0   
POL(active(x1)) = x1   
POL(cons(x1, x2)) = x1   
POL(from(x1)) = 1 + x1   
POL(length(x1)) = 0   
POL(length1(x1)) = 0   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = x1   

The following usable rules [FROCOS05] were oriented: none

(46) Obligation:

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

MARK(from(X)) → ACTIVE(from(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.

(47) 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)))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x0, x1)
ACTIVE(x0, x1)  =  ACTIVE(x0, x1)

Tags:
MARK has argument tags [0,1] and root tag 0
ACTIVE has argument tags [3,0] and root tag 0

Comparison: MIN
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = 1   
POL(MARK(x1)) = 1   
POL(active(x1)) = 0   
POL(cons(x1, x2)) = 1   
POL(from(x1)) = 0   
POL(length(x1)) = 1   
POL(length1(x1)) = 1   
POL(mark(x1)) = 1 + x1   
POL(nil) = 1   
POL(s(x1)) = 1   

The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

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

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.

(49) 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))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x0, x1)
ACTIVE(x0, x1)  =  ACTIVE(x0, x1)

Tags:
MARK has argument tags [1,0] and root tag 0
ACTIVE has argument tags [0,0] and root tag 0

Comparison: MIN
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = 1   
POL(active(x1)) = x1   
POL(cons(x1, x2)) = 1 + x1   
POL(from(x1)) = 1 + x1   
POL(length(x1)) = 0   
POL(length1(x1)) = 0   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = x1   

The following usable rules [FROCOS05] were oriented: none

(50) Obligation:

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

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.

(51) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACTIVE(x0, x1)  =  ACTIVE(x1)
MARK(x0, x1)  =  MARK(x1)

Tags:
ACTIVE has argument tags [1,0] and root tag 0
MARK has argument tags [0,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = 1   
POL(active(x1)) = x1   
POL(cons(x1, x2)) = 1 + x1   
POL(from(x1)) = 1 + x1   
POL(length(x1)) = 0   
POL(length1(x1)) = 0   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = x1   

The following usable rules [FROCOS05] were oriented:

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

(52) Obligation:

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

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

(53) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(s(X)) → ACTIVE(s(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACTIVE(x0, x1)  =  ACTIVE(x0, x1)
MARK(x0, x1)  =  MARK(x0, x1)

Tags:
ACTIVE has argument tags [2,1] and root tag 0
MARK has argument tags [2,3] and root tag 0

Comparison: MIN
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = 0   
POL(active(x1)) = x1   
POL(cons(x1, x2)) = 1   
POL(from(x1)) = 0   
POL(length(x1)) = 1 + x1   
POL(length1(x1)) = 1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = 0   

The following usable rules [FROCOS05] were oriented: none

(54) Obligation:

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

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.

(55) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACTIVE(x0, x1)  =  ACTIVE(x0, x1)
MARK(x0, x1)  =  MARK(x0, x1)

Tags:
ACTIVE has argument tags [0,0] and root tag 0
MARK has argument tags [0,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = 0   
POL(active(x1)) = x1   
POL(cons(x1, x2)) = 1 + x2   
POL(length(x1)) = x1   
POL(length1(x1)) = x1   
POL(mark(x1)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

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

(56) Obligation:

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

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

(57) 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)))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACTIVE(x0, x1)  =  ACTIVE(x1)
MARK(x0, x1)  =  MARK(x1)

Tags:
ACTIVE has argument tags [0,0] and root tag 1
MARK has argument tags [2,0] and root tag 1

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = 1   
POL(active(x1)) = 0   
POL(cons(x1, x2)) = 0   
POL(length(x1)) = 1   
POL(length1(x1)) = 1   
POL(mark(x1)) = 0   
POL(s(x1)) = 0   

The following usable rules [FROCOS05] were oriented: none

(58) Obligation:

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

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.

(59) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(length(X)) → ACTIVE(length(X))
MARK(length1(X)) → ACTIVE(length1(X))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACTIVE(x0, x1)  =  ACTIVE(x1)
MARK(x0, x1)  =  MARK(x0, x1)

Tags:
ACTIVE has argument tags [3,0] and root tag 0
MARK has argument tags [0,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = 1   
POL(MARK(x1)) = 1   
POL(active(x1)) = x1   
POL(length(x1)) = x1   
POL(length1(x1)) = 1 + x1   
POL(mark(x1)) = x1   

The following usable rules [FROCOS05] were oriented: none

(60) Obligation:

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

ACTIVE(length1(X)) → MARK(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.

(61) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(62) TRUE