Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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(after(0, XS)) → MARK(XS)
ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
ACTIVE(after(s(N), cons(X, XS))) → AFTER(N, XS)
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(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))
MARK(after(X1, X2)) → AFTER(mark(X1), mark(X2))
MARK(after(X1, X2)) → MARK(X1)
MARK(after(X1, X2)) → MARK(X2)
MARK(0) → ACTIVE(0)
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)
AFTER(mark(X1), X2) → AFTER(X1, X2)
AFTER(X1, mark(X2)) → AFTER(X1, X2)
AFTER(active(X1), X2) → AFTER(X1, X2)
AFTER(X1, active(X2)) → AFTER(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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 5 SCCs with 9 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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:

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

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:

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

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

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

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

(9) TRUE

(10) 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(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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:

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.

(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:

• 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

(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(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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:

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(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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:

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.

(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:

• 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

(24) TRUE

(25) 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(after(0, XS)) → MARK(XS)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))
MARK(after(X1, X2)) → MARK(X1)
MARK(after(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x₁)) = x₁   
POL(MARK(x₁)) = 1   
POL(active(x₁)) = 0   
POL(after(x₁, x₂)) = 1   
POL(cons(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)
after(X1, active(X2)) → after(X1, X2)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
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)
from(active(X)) → from(X)
from(mark(X)) → from(X)

(27) Obligation:

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

MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(X)) → MARK(X)
ACTIVE(after(0, XS)) → MARK(XS)
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))
MARK(after(X1, X2)) → MARK(X1)
MARK(after(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(after(0, XS)) → MARK(XS)
MARK(after(X1, X2)) → MARK(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(MARK(x1)) =

-I

+

0A

·

x1

POL(from(x1)) =

0A

+

0A

·

x1

POL(ACTIVE(x1)) =

-I

+

0A

·

x1

POL(mark(x1)) =

-I

+

0A

·

x1

POL(cons(x1, x2)) =

-I

+

0A

·

x1

+

0A

·

x2

POL(s(x1)) =

-I

+

0A

·

x1

POL(after(x1, x2)) =

-I

+

0A

·

x1

+

1A

·

x2

POL(0) =

0A

POL(active(x1)) =

-I

+

0A

·

x1

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
active(from(X)) → mark(cons(X, from(s(X))))
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
active(after(0, XS)) → mark(XS)
mark(from(X)) → active(from(mark(X)))
after(X1, active(X2)) → after(X1, X2)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
mark(0) → active(0)
from(active(X)) → from(X)
from(mark(X)) → from(X)

(29) 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)) → MARK(X1)
ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))
MARK(after(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(after(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(MARK(x1)) =

-I

+

0A

·

x1

POL(from(x1)) =

0A

+

0A

·

x1

POL(ACTIVE(x1)) =

-I

+

0A

·

x1

POL(mark(x1)) =

-I

+

0A

·

x1

POL(cons(x1, x2)) =

0A

+

0A

·

x1

+

0A

·

x2

POL(s(x1)) =

-I

+

0A

·

x1

POL(after(x1, x2)) =

0A

+

1A

·

x1

+

0A

·

x2

POL(active(x1)) =

-I

+

0A

·

x1

POL(0) =

0A

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
active(from(X)) → mark(cons(X, from(s(X))))
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
active(after(0, XS)) → mark(XS)
mark(from(X)) → active(from(mark(X)))
after(X1, active(X2)) → after(X1, X2)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
mark(0) → active(0)
from(active(X)) → from(X)
from(mark(X)) → from(X)

(31) 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)) → MARK(X1)
ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(32) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(from(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(MARK(x1)) =

-I

+

0A

·

x1

POL(from(x1)) =

-I

+

1A

·

x1

POL(ACTIVE(x1)) =

-I

+

0A

·

x1

POL(mark(x1)) =

-I

+

0A

·

x1

POL(cons(x1, x2)) =

-I

+

0A

·

x1

+

0A

·

x2

POL(s(x1)) =

-I

+

0A

·

x1

POL(after(x1, x2)) =

0A

+

-I

·

x1

+

0A

·

x2

POL(active(x1)) =

-I

+

0A

·

x1

POL(0) =

0A

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
active(from(X)) → mark(cons(X, from(s(X))))
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
active(after(0, XS)) → mark(XS)
mark(from(X)) → active(from(mark(X)))
after(X1, active(X2)) → after(X1, X2)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
mark(0) → active(0)
from(active(X)) → from(X)
from(mark(X)) → from(X)

(33) 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(cons(X1, X2)) → MARK(X1)
ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(34) 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: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(MARK(x1)) =

-I

+

0A

·

x1

POL(from(x1)) =

0A

+

1A

·

x1

POL(ACTIVE(x1)) =

0A

+

0A

·

x1

POL(mark(x1)) =

-I

+

0A

·

x1

POL(cons(x1, x2)) =

-I

+

1A

·

x1

+

0A

·

x2

POL(s(x1)) =

-I

+

0A

·

x1

POL(after(x1, x2)) =

0A

+

-I

·

x1

+

0A

·

x2

POL(active(x1)) =

-I

+

0A

·

x1

POL(0) =

0A

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
active(from(X)) → mark(cons(X, from(s(X))))
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
active(after(0, XS)) → mark(XS)
mark(from(X)) → active(from(mark(X)))
after(X1, active(X2)) → after(X1, X2)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
mark(0) → active(0)
from(active(X)) → from(X)
from(mark(X)) → from(X)

(35) 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))))
ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(36) 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: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented:

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

(37) Obligation:

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

ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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.

MARK(s(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented:

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

(39) Obligation:

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

ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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.

ACTIVE(from(X)) → MARK(cons(X, from(s(X))))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented:

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

(41) Obligation:

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

ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(42) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
active(from(X)) → mark(cons(X, from(s(X))))
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
active(after(0, XS)) → mark(XS)
mark(from(X)) → active(from(mark(X)))
after(X1, active(X2)) → after(X1, X2)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
mark(0) → active(0)
from(active(X)) → from(X)
from(mark(X)) → from(X)

(43) Obligation:

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

MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
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)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(44) DependencyGraphProof (EQUIVALENT transformation)

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

(45) TRUE