Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

CONS(mark(X1), X2) → CONS(X1, X2)
MARK(2nd(X)) → 2ND(mark(X))
FROM(mark(X)) → FROM(X)
ACTIVE(2nd(cons(X, X1))) → 2ND(cons1(X, X1))
CONS1(active(X1), X2) → CONS1(X1, X2)
MARK(s(X)) → MARK(X)
MARK(from(X)) → FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
FROM(active(X)) → FROM(X)
CONS(X1, mark(X2)) → CONS(X1, X2)
2ND(mark(X)) → 2ND(X)
CONS(X1, active(X2)) → CONS(X1, X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(cons1(X1, X2)) → MARK(X2)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)
MARK(cons1(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(from(X)) → MARK(X)
MARK(s(X)) → S(mark(X))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
CONS1(mark(X1), X2) → CONS1(X1, X2)
MARK(cons1(X1, X2)) → CONS1(mark(X1), mark(X2))
MARK(cons1(X1, X2)) → ACTIVE(cons1(mark(X1), mark(X2)))
ACTIVE(from(X)) → S(X)
ACTIVE(2nd(cons(X, X1))) → CONS1(X, X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(2nd(X)) → MARK(X)
CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
2ND(active(X)) → 2ND(X)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

CONS(mark(X1), X2) → CONS(X1, X2)
MARK(2nd(X)) → 2ND(mark(X))
FROM(mark(X)) → FROM(X)
ACTIVE(2nd(cons(X, X1))) → 2ND(cons1(X, X1))
CONS1(active(X1), X2) → CONS1(X1, X2)
MARK(s(X)) → MARK(X)
MARK(from(X)) → FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
FROM(active(X)) → FROM(X)
CONS(X1, mark(X2)) → CONS(X1, X2)
2ND(mark(X)) → 2ND(X)
CONS(X1, active(X2)) → CONS(X1, X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(cons1(X1, X2)) → MARK(X2)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)
MARK(cons1(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(from(X)) → MARK(X)
MARK(s(X)) → S(mark(X))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
CONS1(mark(X1), X2) → CONS1(X1, X2)
MARK(cons1(X1, X2)) → CONS1(mark(X1), mark(X2))
MARK(cons1(X1, X2)) → ACTIVE(cons1(mark(X1), mark(X2)))
ACTIVE(from(X)) → S(X)
ACTIVE(2nd(cons(X, X1))) → CONS1(X, X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(2nd(X)) → MARK(X)
CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
2ND(active(X)) → 2ND(X)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 6 SCCs with 10 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

S(mark(X)) → S(X)
S(active(X)) → S(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP

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

CONS1(active(X1), X2) → CONS1(X1, X2)
CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)
CONS1(mark(X1), X2) → CONS1(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP

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

CONS1(active(X1), X2) → CONS1(X1, X2)
CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(mark(X1), X2) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP

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

2ND(active(X)) → 2ND(X)
2ND(mark(X)) → 2ND(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP

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

2ND(active(X)) → 2ND(X)
2ND(mark(X)) → 2ND(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof

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

MARK(from(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons1(X1, X2)) → ACTIVE(cons1(mark(X1), mark(X2)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(2nd(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(cons1(X1, X2)) → MARK(X2)
ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(cons1(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons1(X1, X2)) → ACTIVE(cons1(mark(X1), mark(X2)))
MARK(s(X)) → ACTIVE(s(mark(X)))
The remaining pairs can at least be oriented weakly.

MARK(from(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(2nd(X)) → MARK(X)
MARK(cons1(X1, X2)) → MARK(X2)
ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(cons1(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))
Used ordering: Polynomial interpretation [25]:

POL(2nd(x1)) = 1   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = 1   
POL(active(x1)) = 0   
POL(cons(x1, x2)) = 0   
POL(cons1(x1, x2)) = 0   
POL(from(x1)) = 1   
POL(mark(x1)) = 0   
POL(s(x1)) = 0   

The following usable rules [17] 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)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof

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

MARK(from(X)) → MARK(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(2nd(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(cons1(X1, X2)) → MARK(X2)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MARK(2nd(X)) → MARK(X)
ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)
The remaining pairs can at least be oriented weakly.

MARK(from(X)) → MARK(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → MARK(X)
MARK(cons1(X1, X2)) → MARK(X2)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( from(x1) ) =
/0\
\0/
+
/11\
\10/
·x1

M( active(x1) ) =
/0\
\0/
+
/10\
\01/
·x1

M( cons(x1, x2) ) =
/0\
\0/
+
/11\
\00/
·x1+
/00\
\10/
·x2

M( cons1(x1, x2) ) =
/0\
\0/
+
/10\
\01/
·x1+
/10\
\00/
·x2

M( 2nd(x1) ) =
/1\
\0/
+
/11\
\11/
·x1

M( mark(x1) ) =
/0\
\0/
+
/10\
\01/
·x1

M( s(x1) ) =
/0\
\0/
+
/10\
\00/
·x1

Tuple symbols:
M( MARK(x1) ) = 0+
[1,0]
·x1

M( ACTIVE(x1) ) = 0+
[1,0]
·x1


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] 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)
s(active(X)) → s(X)
s(mark(X)) → s(X)
active(from(X)) → mark(cons(X, from(s(X))))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(2nd(X)) → active(2nd(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(from(X)) → active(from(mark(X)))
active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof

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

MARK(from(X)) → MARK(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → MARK(X)
MARK(cons1(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(cons1(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MARK(cons1(X1, X2)) → MARK(X2)
MARK(cons1(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.

MARK(from(X)) → MARK(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( from(x1) ) =
/0\
\0/
+
/11\
\10/
·x1

M( active(x1) ) =
/0\
\0/
+
/10\
\01/
·x1

M( cons(x1, x2) ) =
/0\
\0/
+
/11\
\00/
·x1+
/00\
\10/
·x2

M( cons1(x1, x2) ) =
/1\
\0/
+
/10\
\00/
·x1+
/10\
\10/
·x2

M( 2nd(x1) ) =
/0\
\0/
+
/01\
\01/
·x1

M( mark(x1) ) =
/0\
\0/
+
/10\
\01/
·x1

M( s(x1) ) =
/0\
\0/
+
/10\
\00/
·x1

Tuple symbols:
M( MARK(x1) ) = 0+
[1,0]
·x1

M( ACTIVE(x1) ) = 0+
[1,0]
·x1


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] 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)
s(active(X)) → s(X)
s(mark(X)) → s(X)
active(from(X)) → mark(cons(X, from(s(X))))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(2nd(X)) → active(2nd(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(from(X)) → active(from(mark(X)))
active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ QDPOrderProof

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

MARK(from(X)) → MARK(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → MARK(X)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.

MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → MARK(X)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( from(x1) ) =
/1\
\1/
+
/01\
\11/
·x1

M( active(x1) ) =
/0\
\0/
+
/10\
\01/
·x1

M( cons(x1, x2) ) =
/0\
\1/
+
/00\
\11/
·x1+
/01\
\00/
·x2

M( cons1(x1, x2) ) =
/0\
\0/
+
/00\
\00/
·x1+
/01\
\00/
·x2

M( 2nd(x1) ) =
/0\
\0/
+
/10\
\10/
·x1

M( mark(x1) ) =
/0\
\0/
+
/10\
\01/
·x1

M( s(x1) ) =
/0\
\0/
+
/00\
\01/
·x1

Tuple symbols:
M( MARK(x1) ) = 0+
[0,1]
·x1

M( ACTIVE(x1) ) = 0+
[0,1]
·x1


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] 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)
s(active(X)) → s(X)
s(mark(X)) → s(X)
active(from(X)) → mark(cons(X, from(s(X))))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(2nd(X)) → active(2nd(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(from(X)) → active(from(mark(X)))
active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
QDP
                            ↳ QDPOrderProof

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

MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → MARK(X)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.

ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))
Used ordering: Polynomial interpretation with max and min functions [25]:

POL(2nd(x1)) = 0   
POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = x1   
POL(active(x1)) = 0   
POL(cons(x1, x2)) = 0   
POL(cons1(x1, x2)) = 0   
POL(from(x1)) = 1   
POL(mark(x1)) = 0   
POL(s(x1)) = 1 + x1   

The following usable rules [17] were oriented:

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)
2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
QDP
                                ↳ QDPOrderProof

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

ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
The remaining pairs can at least be oriented weakly.

ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))
Used ordering: Polynomial interpretation with max and min functions [25]:

POL(2nd(x1)) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = 0   
POL(active(x1)) = 0   
POL(cons(x1, x2)) = 0   
POL(cons1(x1, x2)) = 0   
POL(from(x1)) = 1   
POL(mark(x1)) = 0   
POL(s(x1)) = 0   

The following usable rules [17] were oriented:

2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ QDPOrderProof
QDP
                                    ↳ QDPOrderProof

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

ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
The remaining pairs can at least be oriented weakly.

MARK(2nd(X)) → ACTIVE(2nd(mark(X)))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( from(x1) ) =
/0\
\1/
+
/11\
\00/
·x1

M( active(x1) ) =
/0\
\0/
+
/10\
\01/
·x1

M( cons(x1, x2) ) =
/0\
\1/
+
/11\
\00/
·x1+
/10\
\10/
·x2

M( cons1(x1, x2) ) =
/0\
\0/
+
/00\
\00/
·x1+
/10\
\10/
·x2

M( 2nd(x1) ) =
/0\
\0/
+
/01\
\10/
·x1

M( mark(x1) ) =
/0\
\0/
+
/10\
\01/
·x1

M( s(x1) ) =
/0\
\0/
+
/00\
\00/
·x1

Tuple symbols:
M( MARK(x1) ) = 1+
[1,0]
·x1

M( ACTIVE(x1) ) = 1+
[1,0]
·x1


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

from(active(X)) → from(X)
from(mark(X)) → from(X)
active(from(X)) → mark(cons(X, from(s(X))))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(2nd(X)) → active(2nd(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(from(X)) → active(from(mark(X)))
active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
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)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ QDPOrderProof
                                  ↳ QDP
                                    ↳ QDPOrderProof
QDP
                                        ↳ DependencyGraphProof

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

MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.