Termination w.r.t. Q proof of /tmp/tmp0J05Pp/Ex2_Luc03b

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 QDPOrderProof (⇔)

    ↳18 QDP

    ↳19 PisEmptyProof (⇔)

    ↳20 TRUE

  ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 PisEmptyProof (⇔)

    ↳27 TRUE

  ↳28 QDP

    ↳29 QDPOrderProof (⇔)

    ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 QDPOrderProof (⇔)

    ↳34 QDP

    ↳35 PisEmptyProof (⇔)

    ↳36 TRUE

  ↳37 QDP

    ↳38 QDPOrderProof (⇔)

    ↳39 QDP

    ↳40 QDPOrderProof (⇔)

    ↳41 QDP

    ↳42 PisEmptyProof (⇔)

    ↳43 TRUE

  ↳44 QDP

    ↳45 QDPOrderProof (⇔)

    ↳46 QDP

    ↳47 QDPOrderProof (⇔)

    ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 PisEmptyProof (⇔)

    ↳52 TRUE

  ↳53 QDP

    ↳54 QDPOrderProof (⇔)

    ↳55 QDP

    ↳56 QDPOrderProof (⇔)

    ↳57 QDP

    ↳58 DependencyGraphProof (⇔)

    ↳59 QDP

    ↳60 QDPOrderProof (⇔)

    ↳61 QDP

    ↳62 PisEmptyProof (⇔)

    ↳63 TRUE

(0) Obligation:

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

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(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(fst(0, Z)) → MARK(nil)
ACTIVE(fst(s(X), cons(Y, Z))) → MARK(cons(Y, fst(X, Z)))
ACTIVE(fst(s(X), cons(Y, Z))) → CONS(Y, fst(X, Z))
ACTIVE(fst(s(X), cons(Y, Z))) → FST(X, Z)
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(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
ACTIVE(add(s(X), Y)) → S(add(X, Y))
ACTIVE(add(s(X), Y)) → ADD(X, Y)
ACTIVE(len(nil)) → MARK(0)
ACTIVE(len(cons(X, Z))) → MARK(s(len(Z)))
ACTIVE(len(cons(X, Z))) → S(len(Z))
ACTIVE(len(cons(X, Z))) → LEN(Z)
MARK(fst(X1, X2)) → ACTIVE(fst(mark(X1), mark(X2)))
MARK(fst(X1, X2)) → FST(mark(X1), mark(X2))
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(0) → ACTIVE(0)
MARK(nil) → ACTIVE(nil)
MARK(s(X)) → ACTIVE(s(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(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(add(X1, X2)) → ADD(mark(X1), mark(X2))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → ACTIVE(len(mark(X)))
MARK(len(X)) → LEN(mark(X))
MARK(len(X)) → MARK(X)
FST(mark(X1), X2) → FST(X1, X2)
FST(X1, mark(X2)) → FST(X1, X2)
FST(active(X1), X2) → FST(X1, X2)
FST(X1, active(X2)) → FST(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(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)
FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(X)
ADD(mark(X1), X2) → ADD(X1, X2)
ADD(X1, mark(X2)) → ADD(X1, X2)
ADD(active(X1), X2) → ADD(X1, X2)
ADD(X1, active(X2)) → ADD(X1, X2)
LEN(mark(X)) → LEN(X)
LEN(active(X)) → LEN(X)

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(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 7 SCCs with 18 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

LEN(active(X)) → LEN(X)
LEN(mark(X)) → LEN(X)

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LEN(mark(X)) → LEN(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LEN(x1) = LEN(x1)
active(x1) = x1
mark(x1) = mark(x1)
fst(x1, x2) = fst(x1, x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x2)
len(x1) = len(x1)

Recursive Path Order [RPO].
Precedence:

from > [LEN₁, mark₁, fst₂, 0, nil, s, cons, add₁, len₁]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(7) Obligation:

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

LEN(active(X)) → LEN(X)

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LEN(active(X)) → LEN(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LEN(x1) = LEN(x1)
active(x1) = active(x1)
fst(x1, x2) = fst(x1)
0 = 0
mark(x1) = mark(x1)
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len

Recursive Path Order [RPO].
Precedence:

len > 0 > [mark₁, add₂] > active₁ > LEN₁ > [fst₁, nil, cons]
len > [s, from] > [mark₁, add₂] > active₁ > LEN₁ > [fst₁, nil, cons]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(9) Obligation:

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

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

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

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADD(X1, mark(X2)) → ADD(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2) = ADD(x2)
mark(x1) = mark(x1)
active(x1) = x1
fst(x1, x2) = fst(x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x2)
len(x1) = len(x1)

Recursive Path Order [RPO].
Precedence:

from > [ADD₁, mark₁, fst₁, 0, nil, s, cons, add₁, len₁]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(14) Obligation:

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

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

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADD(mark(X1), X2) → ADD(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2) = ADD(x1, x2)
mark(x1) = mark(x1)
active(x1) = x1
fst(x1, x2) = fst(x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len(x1)

Recursive Path Order [RPO].
Precedence:

from > [ADD₂, mark₁, fst₁, 0, nil, s, cons, add₂, len₁]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(16) Obligation:

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

ADD(active(X1), X2) → ADD(X1, X2)
ADD(X1, active(X2)) → ADD(X1, X2)

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADD(active(X1), X2) → ADD(X1, X2)
ADD(X1, active(X2)) → ADD(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2) = ADD(x1, x2)
active(x1) = active(x1)
fst(x1, x2) = fst
0 = 0
mark(x1) = mark(x1)
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len

Recursive Path Order [RPO].
Precedence:

fst > [mark₁, add₂] > 0 > nil > active₁ > [ADD₂, cons]
fst > [mark₁, add₂] > s > active₁ > [ADD₂, cons]
from > [mark₁, add₂] > 0 > nil > active₁ > [ADD₂, cons]
from > [mark₁, add₂] > s > active₁ > [ADD₂, cons]
len > [mark₁, add₂] > 0 > nil > active₁ > [ADD₂, cons]
len > [mark₁, add₂] > s > active₁ > [ADD₂, cons]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(18) Obligation:

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

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE

(21) 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(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(mark(X)) → FROM(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1) = FROM(x1)
active(x1) = x1
mark(x1) = mark(x1)
fst(x1, x2) = fst(x1, x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x2)
len(x1) = len(x1)

Recursive Path Order [RPO].
Precedence:

from > [FROM₁, mark₁, fst₂, 0, nil, s, cons, add₁, len₁]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(23) Obligation:

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

FROM(active(X)) → FROM(X)

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(active(X)) → FROM(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1) = FROM(x1)
active(x1) = active(x1)
fst(x1, x2) = fst(x1)
0 = 0
mark(x1) = mark(x1)
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len

Recursive Path Order [RPO].
Precedence:

len > 0 > [mark₁, add₂] > active₁ > FROM₁ > [fst₁, nil, cons]
len > [s, from] > [mark₁, add₂] > active₁ > FROM₁ > [fst₁, nil, cons]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(25) Obligation:

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

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE

(28) 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(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(X1, mark(X2)) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x2)
mark(x1) = mark(x1)
active(x1) = x1
fst(x1, x2) = fst(x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x2)
len(x1) = len(x1)

Recursive Path Order [RPO].
Precedence:

from > [CONS₁, mark₁, fst₁, 0, nil, s, cons, add₁, len₁]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(30) Obligation:

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

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(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(mark(X1), X2) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1, x2)
mark(x1) = mark(x1)
active(x1) = x1
fst(x1, x2) = fst(x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len(x1)

Recursive Path Order [RPO].
Precedence:

from > [CONS₂, mark₁, fst₁, 0, nil, s, cons, add₂, len₁]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(32) Obligation:

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

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

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1, x2)
active(x1) = active(x1)
fst(x1, x2) = fst
0 = 0
mark(x1) = mark(x1)
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len

Recursive Path Order [RPO].
Precedence:

fst > [mark₁, add₂] > 0 > nil > active₁ > [CONS₂, cons]
fst > [mark₁, add₂] > s > active₁ > [CONS₂, cons]
from > [mark₁, add₂] > 0 > nil > active₁ > [CONS₂, cons]
from > [mark₁, add₂] > s > active₁ > [CONS₂, cons]
len > [mark₁, add₂] > 0 > nil > active₁ > [CONS₂, cons]
len > [mark₁, add₂] > s > active₁ > [CONS₂, cons]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(34) Obligation:

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

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) TRUE

(37) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(38) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(mark(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = S(x1)
active(x1) = x1
mark(x1) = mark(x1)
fst(x1, x2) = fst(x1, x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x2)
len(x1) = len(x1)

Recursive Path Order [RPO].
Precedence:

from > [S₁, mark₁, fst₂, 0, nil, s, cons, add₁, len₁]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(39) Obligation:

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

S(active(X)) → S(X)

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(active(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = S(x1)
active(x1) = active(x1)
fst(x1, x2) = fst(x1)
0 = 0
mark(x1) = mark(x1)
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len

Recursive Path Order [RPO].
Precedence:

len > 0 > [mark₁, add₂] > active₁ > S₁ > [fst₁, nil, cons]
len > [s, from] > [mark₁, add₂] > active₁ > S₁ > [fst₁, nil, cons]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(41) Obligation:

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

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) TRUE

(44) Obligation:

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

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

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FST(X1, mark(X2)) → FST(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FST(x1, x2) = FST(x2)
mark(x1) = mark(x1)
active(x1) = x1
fst(x1, x2) = fst(x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x2)
len(x1) = len(x1)

Recursive Path Order [RPO].
Precedence:

from > [FST₁, mark₁, fst₁, 0, nil, s, cons, add₁, len₁]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(46) Obligation:

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

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

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FST(mark(X1), X2) → FST(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FST(x1, x2) = FST(x1, x2)
mark(x1) = mark(x1)
active(x1) = x1
fst(x1, x2) = fst(x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len(x1)

Recursive Path Order [RPO].
Precedence:

from > [FST₂, mark₁, fst₁, 0, nil, s, cons, add₂, len₁]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(48) Obligation:

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

FST(active(X1), X2) → FST(X1, X2)
FST(X1, active(X2)) → FST(X1, X2)

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FST(active(X1), X2) → FST(X1, X2)
FST(X1, active(X2)) → FST(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FST(x1, x2) = FST(x1, x2)
active(x1) = active(x1)
fst(x1, x2) = fst
0 = 0
mark(x1) = mark(x1)
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len

Recursive Path Order [RPO].
Precedence:

fst > [mark₁, add₂] > 0 > nil > active₁ > [FST₂, cons]
fst > [mark₁, add₂] > s > active₁ > [FST₂, cons]
from > [mark₁, add₂] > 0 > nil > active₁ > [FST₂, cons]
from > [mark₁, add₂] > s > active₁ > [FST₂, cons]
len > [mark₁, add₂] > 0 > nil > active₁ > [FST₂, cons]
len > [mark₁, add₂] > s > active₁ > [FST₂, cons]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(50) Obligation:

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

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(51) PisEmptyProof (EQUIVALENT transformation)

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

(52) TRUE

(53) Obligation:

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

ACTIVE(fst(s(X), cons(Y, Z))) → MARK(cons(Y, fst(X, Z)))
MARK(fst(X1, X2)) → ACTIVE(fst(mark(X1), mark(X2)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(s(X)) → ACTIVE(s(X))
ACTIVE(add(0, X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(len(cons(X, Z))) → MARK(s(len(Z)))
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → ACTIVE(len(mark(X)))
MARK(len(X)) → MARK(X)

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(fst(s(X), cons(Y, Z))) → MARK(cons(Y, fst(X, Z)))
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
fst(x1, x2) = fst(x1, x2)
s(x1) = s
cons(x1, x2) = x1
MARK(x1) = MARK(x1)
mark(x1) = x1
from(x1) = x1
add(x1, x2) = add(x1, x2)
0 = 0
len(x1) = x1
active(x1) = x1
nil = nil

Recursive Path Order [RPO].
Precedence:

[ACTIVE₁, fst₂, s, MARK₁, add₂, 0, nil]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(55) Obligation:

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

MARK(fst(X1, X2)) → ACTIVE(fst(mark(X1), mark(X2)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(s(X)) → ACTIVE(s(X))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(len(cons(X, Z))) → MARK(s(len(Z)))
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(len(X)) → ACTIVE(len(mark(X)))
MARK(len(X)) → MARK(X)

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(56) 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))))
ACTIVE(len(cons(X, Z))) → MARK(s(len(Z)))
MARK(from(X)) → MARK(X)
MARK(len(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = x1
fst(x1, x2) = fst(x2)
ACTIVE(x1) = x1
mark(x1) = x1
from(x1) = from(x1)
cons(x1, x2) = x1
s(x1) = s
len(x1) = len(x1)
add(x1, x2) = add(x2)
active(x1) = x1
0 = 0
nil = nil

Recursive Path Order [RPO].
Precedence:

from₁ > [fst₁, s, len₁, add₁, 0, nil]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(57) Obligation:

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

MARK(fst(X1, X2)) → ACTIVE(fst(mark(X1), mark(X2)))
MARK(s(X)) → ACTIVE(s(X))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(len(X)) → ACTIVE(len(mark(X)))

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(58) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 6 less nodes.

(59) Obligation:

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

MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(60) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = x1
cons(x1, x2) = cons(x1)
active(x1) = x1
fst(x1, x2) = fst(x2)
0 = 0
mark(x1) = x1
nil = nil
s(x1) = s
from(x1) = from(x1)
add(x1, x2) = add(x2)
len(x1) = len

Recursive Path Order [RPO].
Precedence:

from₁ > [cons₁, fst₁, 0, nil, s, add₁, len]

The following usable rules [FROCOS05] were oriented:

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

(61) Obligation:

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

active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)

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

(62) PisEmptyProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) QDPOrderProof (EQUIVALENT transformation)

(7) Obligation:

(8) QDPOrderProof (EQUIVALENT transformation)

(9) Obligation:

(10) PisEmptyProof (EQUIVALENT transformation)

(11) TRUE

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) QDPOrderProof (EQUIVALENT transformation)

(16) Obligation:

(17) QDPOrderProof (EQUIVALENT transformation)

(18) Obligation:

(19) PisEmptyProof (EQUIVALENT transformation)

(20) TRUE

(21) Obligation:

(22) QDPOrderProof (EQUIVALENT transformation)

(23) Obligation:

(24) QDPOrderProof (EQUIVALENT transformation)

(25) Obligation:

(26) PisEmptyProof (EQUIVALENT transformation)

(27) TRUE

(28) Obligation:

(29) QDPOrderProof (EQUIVALENT transformation)

(30) Obligation:

(31) QDPOrderProof (EQUIVALENT transformation)

(32) Obligation:

(33) QDPOrderProof (EQUIVALENT transformation)

(34) Obligation:

(35) PisEmptyProof (EQUIVALENT transformation)

(36) TRUE

(37) Obligation:

(38) QDPOrderProof (EQUIVALENT transformation)

(39) Obligation:

(40) QDPOrderProof (EQUIVALENT transformation)

(41) Obligation:

(42) PisEmptyProof (EQUIVALENT transformation)

(43) TRUE

(44) Obligation:

(45) QDPOrderProof (EQUIVALENT transformation)

(46) Obligation:

(47) QDPOrderProof (EQUIVALENT transformation)

(48) Obligation:

(49) QDPOrderProof (EQUIVALENT transformation)

(50) Obligation:

(51) PisEmptyProof (EQUIVALENT transformation)

(52) TRUE

(53) Obligation:

(54) QDPOrderProof (EQUIVALENT transformation)

(55) Obligation:

(56) QDPOrderProof (EQUIVALENT transformation)

(57) Obligation:

(58) DependencyGraphProof (EQUIVALENT transformation)

(59) Obligation:

(60) QDPOrderProof (EQUIVALENT transformation)

(61) Obligation:

(62) PisEmptyProof (EQUIVALENT transformation)

(63) TRUE