(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)  =  x1
active(x1)  =  x1
mark(x1)  =  mark(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(x1)

Lexicographic Path Order [LPO].
Precedence:
[mark1, fst1, from1, add1, len1] > [0, nil, s]


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) 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.

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

Lexicographic Path Order [LPO].
Precedence:
ADD1 > [nil, s, cons]
len > [mark1, fst1, from1, add1] > 0 > [nil, s, 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)

(10) 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.

(11) 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)
mark(x1)  =  mark(x1)
active(x1)  =  x1
fst(x1, x2)  =  fst(x1)
0  =  0
nil  =  nil
s(x1)  =  s
cons(x1, x2)  =  cons
from(x1)  =  from(x1)
add(x1, x2)  =  add(x2)
len(x1)  =  len(x1)

Lexicographic Path Order [LPO].
Precedence:
[mark1, fst1, from1, add1, len1] > ADD1 > [s, cons]
[mark1, fst1, from1, add1, len1] > [0, nil] > [s, 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)

(12) 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.

(13) 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.

(14) 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)  =  x1
active(x1)  =  x1
mark(x1)  =  mark(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(x1)

Lexicographic Path Order [LPO].
Precedence:
[mark1, fst1, from1, add1, len1] > [0, nil, s]


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)

(15) 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.

(16) 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.

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

Lexicographic Path Order [LPO].
Precedence:
CONS1 > [nil, s, cons]
len > [mark1, fst1, from1, add1] > 0 > [nil, s, 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:
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.

(19) 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
0  =  0
nil  =  nil
s(x1)  =  s
cons(x1, x2)  =  cons
from(x1)  =  from
add(x1, x2)  =  add(x2)
len(x1)  =  len

Lexicographic Path Order [LPO].
Precedence:
fst > [mark1, add1] > CONS2 > s
fst > [mark1, add1] > [0, nil] > s
fst > [mark1, add1] > cons > s
from > [mark1, add1] > CONS2 > s
from > [mark1, add1] > [0, nil] > s
from > [mark1, add1] > cons > s
len > [mark1, add1] > CONS2 > s
len > [mark1, add1] > [0, nil] > s
len > [mark1, add1] > cons > s


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:
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.

(21) 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.

(22) 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)  =  x1
active(x1)  =  x1
mark(x1)  =  mark(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(x1)

Lexicographic Path Order [LPO].
Precedence:
[mark1, fst1, from1, add1, len1] > [0, nil, s]


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:

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.

(24) 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.

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

Lexicographic Path Order [LPO].
Precedence:
FST1 > [nil, s, cons]
len > [mark1, fst1, from1, add1] > 0 > [nil, s, 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)

(26) 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.

(27) 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
0  =  0
nil  =  nil
s(x1)  =  s
cons(x1, x2)  =  cons
from(x1)  =  from
add(x1, x2)  =  add(x2)
len(x1)  =  len

Lexicographic Path Order [LPO].
Precedence:
fst > [mark1, add1] > FST2 > s
fst > [mark1, add1] > [0, nil] > s
fst > [mark1, add1] > cons > s
from > [mark1, add1] > FST2 > s
from > [mark1, add1] > [0, nil] > s
from > [mark1, add1] > cons > s
len > [mark1, add1] > FST2 > s
len > [mark1, add1] > [0, nil] > s
len > [mark1, add1] > cons > s


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)

(28) 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.

(29) 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.

(30) 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)))
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)
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)
MARK(len(X)) → MARK(X)
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)  =  len(x1)
active(x1)  =  x1
nil  =  nil

Lexicographic Path Order [LPO].
Precedence:
from1 > [ACTIVE1, cons1, MARK1] > s > [fst2, nil] > 0
add2 > [ACTIVE1, cons1, MARK1] > s > [fst2, nil] > 0
len1 > [ACTIVE1, cons1, MARK1] > s > [fst2, nil] > 0


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)

(31) 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(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.

(32) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 6 less nodes.

(33) TRUE