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 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 QDPSizeChangeProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 UsableRulesProof (⇔)
    ↳12 QDP
    ↳13 QDPSizeChangeProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 UsableRulesProof (⇔)
    ↳17 QDP
    ↳18 QDPSizeChangeProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 UsableRulesProof (⇔)
    ↳22 QDP
    ↳23 QDPSizeChangeProof (⇔)
    ↳24 TRUE
  ↳25 QDP
    ↳26 UsableRulesProof (⇔)
    ↳27 QDP
    ↳28 QDPSizeChangeProof (⇔)
    ↳29 TRUE
  ↳30 QDP
    ↳31 QDPOrderProof (⇔)
    ↳32 QDP
    ↳33 QDPOrderProof (⇔)
    ↳34 QDP
    ↳35 QDPOrderProof (⇔)
    ↳36 QDP
    ↳37 QDPOrderProof (⇔)
    ↳38 QDP
    ↳39 QDPOrderProof (⇔)
    ↳40 QDP
    ↳41 QDPOrderProof (⇔)
    ↳42 QDP
    ↳43 QDPOrderProof (⇔)
    ↳44 QDP
    ↳45 DependencyGraphProof (⇔)
    ↳46 QDP
    ↳47 UsableRulesProof (⇔)
    ↳48 QDP
    ↳49 QDPSizeChangeProof (⇔)
    ↳50 TRUE

(0) Obligation:

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.

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

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.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 10 less nodes.

(4) Complex Obligation (AND)

(5) 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(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.

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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.

(8) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • S(active(X)) → S(X)
    The graph contains the following edges 1 > 1

  • S(mark(X)) → S(X)
    The graph contains the following edges 1 > 1

(9) TRUE

(10) 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(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.

(11) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(12) Obligation:

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

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

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

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • FROM(active(X)) → FROM(X)
    The graph contains the following edges 1 > 1

  • FROM(mark(X)) → FROM(X)
    The graph contains the following edges 1 > 1

(14) TRUE

(15) 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(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.

(16) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(17) 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)

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

(18) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • CONS(X1, mark(X2)) → CONS(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

  • CONS(mark(X1), X2) → CONS(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • CONS(active(X1), X2) → CONS(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • CONS(X1, active(X2)) → CONS(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

(19) TRUE

(20) Obligation:

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

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

(21) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(22) Obligation:

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

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

(23) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • CONS1(X1, mark(X2)) → CONS1(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

  • CONS1(mark(X1), X2) → CONS1(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • CONS1(active(X1), X2) → CONS1(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • CONS1(X1, active(X2)) → CONS1(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

(24) TRUE

(25) Obligation:

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.

(26) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(27) Obligation:

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.

(28) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • 2ND(active(X)) → 2ND(X)
    The graph contains the following edges 1 > 1

  • 2ND(mark(X)) → 2ND(X)
    The graph contains the following edges 1 > 1

(29) TRUE

(30) Obligation:

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

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

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons1(X1, X2)) → ACTIVE(cons1(mark(X1), mark(X2)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → ACTIVE(s(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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 [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
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)
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)

(32) Obligation:

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

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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.
Used ordering: Matrix interpretation [MATRO]:

POL(MARK(x1)) =
/0\
\1/
 +
/01\
\00/
·x1

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

POL(ACTIVE(x1)) =
/0\
\1/
 +
/01\
\00/
·x1

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

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

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

POL(from(x1)) =
/1\
\0/
 +
/01\
\11/
·x1

POL(s(x1)) =
/0\
\0/
 +
/00\
\01/
·x1

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

The following usable rules [FROCOS05] were oriented:

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)))
s(active(X)) → s(X)
s(mark(X)) → s(X)
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)
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)

(34) Obligation:

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

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

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(MARK(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

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

POL(ACTIVE(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

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

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

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

POL(from(x1)) =
/0\
\1/
 +
/11\
\10/
·x1

POL(s(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

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

The following usable rules [FROCOS05] were oriented:

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)))
s(active(X)) → s(X)
s(mark(X)) → s(X)
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)
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)

(36) Obligation:

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

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

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(2nd(X)) → ACTIVE(2nd(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(2nd(x1)) = 0   
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 [FROCOS05] were oriented:

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

(38) Obligation:

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

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

(39) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons1(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(MARK(x1)) =
/0\
\0/
 +
/01\
\01/
·x1

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

POL(cons1(x1, x2)) =
/0\
\1/
 +
/00\
\11/
·x1 +
/00\
\01/
·x2

POL(ACTIVE(x1)) =
/0\
\0/
 +
/01\
\01/
·x1

POL(from(x1)) =
/1\
\0/
 +
/01\
\11/
·x1

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

POL(s(x1)) =
/0\
\0/
 +
/00\
\01/
·x1

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

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

The following usable rules [FROCOS05] were oriented:

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)))
s(active(X)) → s(X)
s(mark(X)) → s(X)
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)
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)

(40) Obligation:

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

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(2nd(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(MARK(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

POL(2nd(x1)) =
/1\
\0/
 +
/11\
\01/
·x1

POL(ACTIVE(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

POL(from(x1)) =
/0\
\0/
 +
/11\
\10/
·x1

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

POL(s(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

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

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

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

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
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)
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)

(42) Obligation:

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

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

(43) 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: Matrix interpretation [MATRO]:

POL(ACTIVE(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

POL(from(x1)) =
/1\
\1/
 +
/11\
\10/
·x1

POL(MARK(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

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

POL(s(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

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

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

POL(cons1(x1, x2)) =
/0\
\0/
 +
/11\
\00/
·x1 +
/10\
\00/
·x2

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

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
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)
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)

(44) Obligation:

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

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

(45) DependencyGraphProof (EQUIVALENT transformation)

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

(46) 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:

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.

(47) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(48) Obligation:

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

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

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

(49) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • MARK(s(X)) → MARK(X)
    The graph contains the following edges 1 > 1

  • MARK(cons(X1, X2)) → MARK(X1)
    The graph contains the following edges 1 > 1

(50) TRUE