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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 PisEmptyProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 QDPOrderProof (⇔)
    ↳23 QDP
    ↳24 QDPOrderProof (⇔)
    ↳25 QDP
    ↳26 PisEmptyProof (⇔)
    ↳27 TRUE
  ↳28 QDP
    ↳29 QDPOrderProof (⇔)
    ↳30 QDP
    ↳31 QDPOrderProof (⇔)
    ↳32 QDP
    ↳33 QDPOrderProof (⇔)
    ↳34 QDP
    ↳35 PisEmptyProof (⇔)
    ↳36 TRUE
  ↳37 QDP
    ↳38 QDPOrderProof (⇔)
    ↳39 QDP
    ↳40 QDPOrderProof (⇔)
    ↳41 QDP
    ↳42 PisEmptyProof (⇔)
    ↳43 TRUE
  ↳44 QDP
    ↳45 QDPOrderProof (⇔)
    ↳46 QDP
    ↳47 QDPOrderProof (⇔)
    ↳48 QDP
    ↳49 QDPOrderProof (⇔)
    ↳50 QDP
    ↳51 PisEmptyProof (⇔)
    ↳52 TRUE
  ↳53 QDP
    ↳54 QDPOrderProof (⇔)
    ↳55 QDP
    ↳56 DependencyGraphProof (⇔)
    ↳57 QDP
    ↳58 QDPOrderProof (⇔)
    ↳59 QDP
    ↳60 PisEmptyProof (⇔)
    ↳61 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(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)  =  x1
active(x1)  =  active(x1)
mark(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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.

(8) 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
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

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

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ADD(X1, 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
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(14) 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)

The TRS R consists of the following rules:

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

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ADD(mark(X1), X2) → ADD(X1, X2)
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)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

ADD(X1, mark(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, 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)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

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

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE

(21) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

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)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

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

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

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE

(28) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(X1, 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
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(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)

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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(32) Obligation:

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

CONS(X1, mark(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)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(34) Obligation:

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

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

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

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) TRUE

(37) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

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

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

(38) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(39) Obligation:

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

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.

(40) 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
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(41) Obligation:

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

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

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

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) TRUE

(44) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FST(X1, 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
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(46) 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)

The TRS R consists of the following rules:

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

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

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FST(mark(X1), X2) → FST(X1, X2)
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)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

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

FST(X1, mark(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(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)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(50) Obligation:

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

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

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

(51) PisEmptyProof (EQUIVALENT transformation)

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

(52) TRUE

(53) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(fst(s(X), cons(Y, Z))) → MARK(cons(Y, fst(X, Z)))
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)))
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)  =  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, fst2, s, MARK1, add2, 0, len1, nil]


The following usable rules [FROCOS05] were oriented:

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

(55) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(56) DependencyGraphProof (EQUIVALENT transformation)

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

(57) Obligation:

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

MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

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

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

(58) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
[MARK1, cons2]


The following usable rules [FROCOS05] were oriented: none

(59) 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.

(60) PisEmptyProof (EQUIVALENT transformation)

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

(61) TRUE