Termination w.r.t. Q proof of /tmp/tmpKMdx7a/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 QDPOrderProof (⇔)

    ↳20 QDP

    ↳21 PisEmptyProof (⇔)

    ↳22 TRUE

  ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 PisEmptyProof (⇔)

    ↳29 TRUE

  ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 QDPOrderProof (⇔)

    ↳34 QDP

    ↳35 QDPOrderProof (⇔)

    ↳36 QDP

    ↳37 QDPOrderProof (⇔)

    ↳38 QDP

    ↳39 PisEmptyProof (⇔)

    ↳40 TRUE

  ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 PisEmptyProof (⇔)

    ↳47 TRUE

  ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 QDPOrderProof (⇔)

    ↳52 QDP

    ↳53 QDPOrderProof (⇔)

    ↳54 QDP

    ↳55 QDPOrderProof (⇔)

    ↳56 QDP

    ↳57 PisEmptyProof (⇔)

    ↳58 TRUE

  ↳59 QDP

    ↳60 QDPOrderProof (⇔)

    ↳61 QDP

    ↳62 QDPOrderProof (⇔)

    ↳63 QDP

    ↳64 QDPOrderProof (⇔)

    ↳65 QDP

    ↳66 QDPOrderProof (⇔)

    ↳67 QDP

    ↳68 PisEmptyProof (⇔)

    ↳69 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(x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len

Lexicographic path order with status [LPO].
Quasi-Precedence:

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

Status:

LEN₁: [1]
mark₁: [1]
fst₁: [1]
0: []
nil: []
s: []
cons: []
from: []
add₂: [2,1]
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)

(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
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

Lexicographic path order with status [LPO].
Quasi-Precedence:

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

Status:

LEN₁: [1]
active₁: [1]
fst: []
0: []
mark₁: [1]
nil: []
s: []
cons: []
from: []
add₂: [2,1]
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)

(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(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) = x1
mark(x1) = mark(x1)
active(x1) = x1
fst(x1, x2) = fst
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

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

Status:

mark₁: [1]
fst: []
0: []
nil: []
s: []
cons: []
from: []
add₂: [2,1]
len₁: [1]

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(X1, mark(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(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) = x2
mark(x1) = mark(x1)
active(x1) = x1
fst(x1, x2) = fst(x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = x1
from(x1) = from(x1)
add(x1, x2) = add(x2)
len(x1) = len

Lexicographic path order with status [LPO].
Quasi-Precedence:

len > [mark₁, fst₁, 0, from₁, add₁] > [nil, s]

Status:

mark₁: [1]
fst₁: [1]
0: []
nil: []
s: []
from₁: [1]
add₁: [1]
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)

(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(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) = 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(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

fst > nil > [mark₁, add₂] > active₁ > 0 > s
fst > nil > [mark₁, add₂] > active₁ > len₁ > s
fst > cons > [mark₁, add₂] > active₁ > 0 > s
fst > cons > [mark₁, add₂] > active₁ > len₁ > s
from > cons > [mark₁, add₂] > active₁ > 0 > s
from > cons > [mark₁, add₂] > active₁ > len₁ > s

Status:

active₁: [1]
fst: []
0: []
mark₁: [1]
nil: []
s: []
cons: []
from: []
add₂: [2,1]
len₁: [1]

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:
The TRS P consists of the following rules:

ADD(active(X1), 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.

(19) 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)

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(x1, x2)
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

Lexicographic path order with status [LPO].
Quasi-Precedence:

from > [fst₂, 0, mark₁, add₂] > nil > active₁ > ADD₂ > [s, cons]
len > [fst₂, 0, mark₁, add₂] > nil > active₁ > ADD₂ > [s, cons]

Status:

ADD₂: [2,1]
active₁: [1]
fst₂: [1,2]
0: []
mark₁: [1]
nil: []
s: []
cons: []
from: []
add₂: [2,1]
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)

(20) 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.

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) 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.

(24) 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(x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len

Lexicographic path order with status [LPO].
Quasi-Precedence:

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

Status:

FROM₁: [1]
mark₁: [1]
fst₁: [1]
0: []
nil: []
s: []
cons: []
from: []
add₂: [2,1]
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)

(25) 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.

(26) 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
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

Lexicographic path order with status [LPO].
Quasi-Precedence:

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

Status:

FROM₁: [1]
active₁: [1]
fst: []
0: []
mark₁: [1]
nil: []
s: []
cons: []
from: []
add₂: [2,1]
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)

(27) 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.

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE

(30) 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.

(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) = x1
mark(x1) = mark(x1)
active(x1) = x1
fst(x1, x2) = fst
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

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

Status:

mark₁: [1]
fst: []
0: []
nil: []
s: []
cons: []
from: []
add₂: [2,1]
len₁: [1]

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

The TRS R consists of the following rules:

active(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(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) = x2
mark(x1) = mark(x1)
active(x1) = x1
fst(x1, x2) = fst(x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = x1
from(x1) = from(x1)
add(x1, x2) = add(x2)
len(x1) = len

Lexicographic path order with status [LPO].
Quasi-Precedence:

len > [mark₁, fst₁, 0, from₁, add₁] > [nil, s]

Status:

mark₁: [1]
fst₁: [1]
0: []
nil: []
s: []
from₁: [1]
add₁: [1]
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)

(34) 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.

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(active(X1), X2) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1)
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

Lexicographic path order with status [LPO].
Quasi-Precedence:

fst > [mark₁, add₂] > cons > [CONS₁, active₁] > [0, nil, s]
from > [mark₁, add₂] > cons > [CONS₁, active₁] > [0, nil, s]
len > [mark₁, add₂] > cons > [CONS₁, active₁] > [0, nil, s]

Status:

CONS₁: [1]
active₁: [1]
fst: []
0: []
mark₁: [1]
nil: []
s: []
cons: []
from: []
add₂: [2,1]
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)

(36) Obligation:

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

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.

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(X1, active(X2)) → CONS(X1, X2)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

from > [fst₂, mark₁, add₂] > [active₁, 0, nil, s, cons]
len > [fst₂, mark₁, add₂] > [active₁, 0, nil, s, cons]

Status:

active₁: [1]
fst₂: [1,2]
0: []
mark₁: [1]
nil: []
s: []
cons: []
from: []
add₂: [2,1]
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)

(38) 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.

(39) PisEmptyProof (EQUIVALENT transformation)

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

(40) TRUE

(41) 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.

(42) 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(x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len

Lexicographic path order with status [LPO].
Quasi-Precedence:

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

Status:

S₁: [1]
mark₁: [1]
fst₁: [1]
0: []
nil: []
s: []
cons: []
from: []
add₂: [2,1]
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)

(43) 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.

(44) 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
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

Lexicographic path order with status [LPO].
Quasi-Precedence:

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

Status:

S₁: [1]
active₁: [1]
fst: []
0: []
mark₁: [1]
nil: []
s: []
cons: []
from: []
add₂: [2,1]
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)

(45) 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.

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) 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.

(49) 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) = x1
mark(x1) = mark(x1)
active(x1) = x1
fst(x1, x2) = fst
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = cons
from(x1) = from
add(x1, x2) = add(x1, x2)
len(x1) = len(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

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

Status:

mark₁: [1]
fst: []
0: []
nil: []
s: []
cons: []
from: []
add₂: [2,1]
len₁: [1]

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:
The TRS P consists of the following rules:

FST(X1, mark(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.

(51) 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) = x2
mark(x1) = mark(x1)
active(x1) = x1
fst(x1, x2) = fst(x2)
0 = 0
nil = nil
s(x1) = s
cons(x1, x2) = x1
from(x1) = from(x1)
add(x1, x2) = add(x2)
len(x1) = len

Lexicographic path order with status [LPO].
Quasi-Precedence:

len > [mark₁, fst₁, 0, from₁, add₁] > [nil, s]

Status:

mark₁: [1]
fst₁: [1]
0: []
nil: []
s: []
from₁: [1]
add₁: [1]
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)

(52) 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.

(53) 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)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FST(x1, x2) = FST(x1)
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

Lexicographic path order with status [LPO].
Quasi-Precedence:

fst > [mark₁, add₂] > cons > [FST₁, active₁] > [0, nil, s]
from > [mark₁, add₂] > cons > [FST₁, active₁] > [0, nil, s]
len > [mark₁, add₂] > cons > [FST₁, active₁] > [0, nil, s]

Status:

FST₁: [1]
active₁: [1]
fst: []
0: []
mark₁: [1]
nil: []
s: []
cons: []
from: []
add₂: [2,1]
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)

(54) Obligation:

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

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.

(55) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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) = x2
active(x1) = active(x1)
fst(x1, x2) = fst(x1, x2)
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

Lexicographic path order with status [LPO].
Quasi-Precedence:

from > [fst₂, mark₁, add₂] > [active₁, 0, nil, s, cons]
len > [fst₂, mark₁, add₂] > [active₁, 0, nil, s, cons]

Status:

active₁: [1]
fst₂: [1,2]
0: []
mark₁: [1]
nil: []
s: []
cons: []
from: []
add₂: [2,1]
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)

(56) 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.

(57) PisEmptyProof (EQUIVALENT transformation)

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

(58) TRUE

(59) 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.

(60) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(s(X)) → ACTIVE(s(X))

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[fst, cons, MARK, mark, from, add, len, nil] > 0 > s

Status:

fst: []
s: []
cons: []
MARK: []
mark: []
from: []
add: []
0: []
len: []
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)

(61) 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)
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.

(62) 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))

[fst, MARK, mark, from, add, 0, len, nil] > s > cons

Status:

fst: []
s: []
cons: []
MARK: []
mark: []
from: []
add: []
0: []
len: []
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)

(63) 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)
ACTIVE(add(0, X)) → MARK(X)
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.

(64) 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(cons(X1, X2)) → MARK(X1)
ACTIVE(len(cons(X, Z))) → MARK(s(len(Z)))
MARK(from(X)) → MARK(X)
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) = cons(x1)
MARK(x1) = MARK(x1)
mark(x1) = x1
from(x1) = from(x1)
add(x1, x2) = add(x1, x2)
0 = 0
len(x1) = x1
active(x1) = x1
nil = nil

Lexicographic path order with status [LPO].
Quasi-Precedence:

fst₂ > [ACTIVE₁, cons₁, MARK₁, from₁, add₂, 0, nil] > s

Status:

ACTIVE₁: [1]
fst₂: [1,2]
s: []
cons₁: [1]
MARK₁: [1]
from₁: [1]
add₂: [1,2]
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)

(65) 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(from(X)) → ACTIVE(from(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.

(66) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(fst(X1, X2)) → ACTIVE(fst(mark(X1), mark(X2)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(X)) → ACTIVE(from(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 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(x1, x2)
ACTIVE(x1) = ACTIVE
mark(x1) = x1
from(x1) = from
cons(x1, x2) = cons
s(x1) = s
add(x1, x2) = add(x1, x2)
len(x1) = len(x1)
active(x1) = x1
0 = 0
nil = nil

Lexicographic path order with status [LPO].
Quasi-Precedence:

fst₂ > ACTIVE > cons
fst₂ > nil > 0
from > ACTIVE > cons
from > s > cons
add₂ > ACTIVE > cons
add₂ > s > cons
len₁ > ACTIVE > cons
len₁ > s > cons
len₁ > 0

Status:

fst₂: [2,1]
ACTIVE: []
from: []
cons: []
s: []
add₂: [2,1]
len₁: [1]
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)

(67) 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.

(68) 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) QDPOrderProof (EQUIVALENT transformation)

(20) Obligation:

(21) PisEmptyProof (EQUIVALENT transformation)

(22) TRUE

(23) Obligation:

(24) QDPOrderProof (EQUIVALENT transformation)

(25) Obligation:

(26) QDPOrderProof (EQUIVALENT transformation)

(27) Obligation:

(28) PisEmptyProof (EQUIVALENT transformation)

(29) TRUE

(30) Obligation:

(31) QDPOrderProof (EQUIVALENT transformation)

(32) Obligation:

(33) QDPOrderProof (EQUIVALENT transformation)

(34) Obligation:

(35) QDPOrderProof (EQUIVALENT transformation)

(36) Obligation:

(37) QDPOrderProof (EQUIVALENT transformation)

(38) Obligation:

(39) PisEmptyProof (EQUIVALENT transformation)

(40) TRUE

(41) Obligation:

(42) QDPOrderProof (EQUIVALENT transformation)

(43) Obligation:

(44) QDPOrderProof (EQUIVALENT transformation)

(45) Obligation:

(46) PisEmptyProof (EQUIVALENT transformation)

(47) TRUE

(48) Obligation:

(49) QDPOrderProof (EQUIVALENT transformation)

(50) Obligation:

(51) QDPOrderProof (EQUIVALENT transformation)

(52) Obligation:

(53) QDPOrderProof (EQUIVALENT transformation)

(54) Obligation:

(55) QDPOrderProof (EQUIVALENT transformation)

(56) Obligation:

(57) PisEmptyProof (EQUIVALENT transformation)

(58) TRUE

(59) Obligation:

(60) QDPOrderProof (EQUIVALENT transformation)

(61) Obligation:

(62) QDPOrderProof (EQUIVALENT transformation)

(63) Obligation:

(64) QDPOrderProof (EQUIVALENT transformation)

(65) Obligation:

(66) QDPOrderProof (EQUIVALENT transformation)

(67) Obligation:

(68) PisEmptyProof (EQUIVALENT transformation)

(69) TRUE