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 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 QDPOrderProof (⇔)
    ↳21 QDP
    ↳22 QDPOrderProof (⇔)
    ↳23 QDP
    ↳24 PisEmptyProof (⇔)
    ↳25 TRUE
  ↳26 QDP
    ↳27 QDPOrderProof (⇔)
    ↳28 QDP
    ↳29 QDPOrderProof (⇔)
    ↳30 QDP
    ↳31 PisEmptyProof (⇔)
    ↳32 TRUE
  ↳33 QDP
    ↳34 QDPOrderProof (⇔)
    ↳35 QDP
    ↳36 QDPOrderProof (⇔)
    ↳37 QDP
    ↳38 QDPOrderProof (⇔)
    ↳39 QDP
    ↳40 PisEmptyProof (⇔)
    ↳41 TRUE
  ↳42 QDP
    ↳43 QDPOrderProof (⇔)
    ↳44 QDP
    ↳45 QDPOrderProof (⇔)
    ↳46 QDP
    ↳47 PisEmptyProof (⇔)
    ↳48 TRUE
  ↳49 QDP
    ↳50 QDPOrderProof (⇔)
    ↳51 QDP
    ↳52 QDPOrderProof (⇔)
    ↳53 QDP
    ↳54 QDPOrderProof (⇔)
    ↳55 QDP
    ↳56 QDPOrderProof (⇔)
    ↳57 QDP
    ↳58 PisEmptyProof (⇔)
    ↳59 TRUE
  ↳60 QDP
    ↳61 QDPOrderProof (⇔)
    ↳62 QDP
    ↳63 QDPOrderProof (⇔)
    ↳64 QDP
    ↳65 QDPOrderProof (⇔)
    ↳66 QDP
    ↳67 QDPOrderProof (⇔)
    ↳68 QDP
    ↳69 PisEmptyProof (⇔)
    ↳70 TRUE
  ↳71 QDP
    ↳72 QDPOrderProof (⇔)
    ↳73 QDP
    ↳74 QDPOrderProof (⇔)
    ↳75 QDP
    ↳76 PisEmptyProof (⇔)
    ↳77 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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(s(X), cons(Y, Z))) → CONS(Y, fst(X, Z))
ACTIVE(fst(s(X), cons(Y, Z))) → FST(X, Z)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(add(s(X), Y)) → S(add(X, Y))
ACTIVE(add(s(X), Y)) → ADD(X, Y)
ACTIVE(len(cons(X, Z))) → S(len(Z))
ACTIVE(len(cons(X, Z))) → LEN(Z)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(fst(X1, X2)) → FST(active(X1), X2)
ACTIVE(fst(X1, X2)) → ACTIVE(X1)
ACTIVE(fst(X1, X2)) → FST(X1, active(X2))
ACTIVE(fst(X1, X2)) → ACTIVE(X2)
ACTIVE(from(X)) → FROM(active(X))
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(add(X1, X2)) → ADD(active(X1), X2)
ACTIVE(add(X1, X2)) → ACTIVE(X1)
ACTIVE(add(X1, X2)) → ADD(X1, active(X2))
ACTIVE(add(X1, X2)) → ACTIVE(X2)
ACTIVE(len(X)) → LEN(active(X))
ACTIVE(len(X)) → ACTIVE(X)
CONS(mark(X1), X2) → CONS(X1, X2)
FST(mark(X1), X2) → FST(X1, X2)
FST(X1, mark(X2)) → FST(X1, X2)
FROM(mark(X)) → FROM(X)
ADD(mark(X1), X2) → ADD(X1, X2)
ADD(X1, mark(X2)) → ADD(X1, X2)
LEN(mark(X)) → LEN(X)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(fst(X1, X2)) → FST(proper(X1), proper(X2))
PROPER(fst(X1, X2)) → PROPER(X1)
PROPER(fst(X1, X2)) → PROPER(X2)
PROPER(from(X)) → FROM(proper(X))
PROPER(from(X)) → PROPER(X)
PROPER(add(X1, X2)) → ADD(proper(X1), proper(X2))
PROPER(add(X1, X2)) → PROPER(X1)
PROPER(add(X1, X2)) → PROPER(X2)
PROPER(len(X)) → LEN(proper(X))
PROPER(len(X)) → PROPER(X)
S(ok(X)) → S(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
FST(ok(X1), ok(X2)) → FST(X1, X2)
FROM(ok(X)) → FROM(X)
ADD(ok(X1), ok(X2)) → ADD(X1, X2)
LEN(ok(X)) → LEN(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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 9 SCCs with 24 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

S(ok(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


S(ok(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
ok(x1)  =  ok(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(7) 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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

LEN(ok(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


LEN(ok(X)) → LEN(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LEN(x1)  =  LEN(x1)
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
ok1 > LEN1

Status:
LEN1: [1]
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(12) 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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


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 with status [LPO].
Precedence:
trivial

Status:
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(14) 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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) 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(ok(X1), ok(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(18) 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)
ok(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
mark1 > ADD1

Status:
mark1: [1]
ADD1: [1]

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

ADD(X1, mark(X2)) → ADD(X1, X2)
ADD(ok(X1), ok(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ADD(ok(X1), ok(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)  =  x1
ok(x1)  =  ok(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
ok1: [1]
ADD1: [1]

The following usable rules [FROCOS05] were oriented: none

(21) 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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


ADD(X1, mark(X2)) → ADD(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
mark1 > ADD2

Status:
mark1: [1]
ADD2: [1,2]

The following usable rules [FROCOS05] were oriented: none

(23) 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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) Obligation:

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

FROM(ok(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


FROM(ok(X)) → FROM(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1)  =  FROM(x1)
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
ok1 > FROM1

Status:
ok1: [1]
FROM1: [1]

The following usable rules [FROCOS05] were oriented: none

(28) 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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


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 with status [LPO].
Precedence:
trivial

Status:
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(30) 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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) 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(ok(X1), ok(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)
mark(x1)  =  mark(x1)
ok(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
mark1 > FST1

Status:
FST1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

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

FST(X1, mark(X2)) → FST(X1, X2)
FST(ok(X1), ok(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
FST1: [1]
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(37) 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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


FST(X1, mark(X2)) → FST(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
mark1 > FST2

Status:
FST2: [1,2]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(39) 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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(40) PisEmptyProof (EQUIVALENT transformation)

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

(41) TRUE

(42) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)
CONS(mark(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(43) 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)
ok(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic path order with status [LPO].
Precedence:
mark1 > CONS1

Status:
CONS1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(44) Obligation:

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

CONS(ok(X1), ok(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


CONS(ok(X1), ok(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)
ok(x1)  =  ok(x1)

Lexicographic path order with status [LPO].
Precedence:
ok1 > CONS1

Status:
CONS1: [1]
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(46) 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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(47) PisEmptyProof (EQUIVALENT transformation)

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

(48) TRUE

(49) Obligation:

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

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(fst(X1, X2)) → PROPER(X1)
PROPER(fst(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)
PROPER(add(X1, X2)) → PROPER(X1)
PROPER(add(X1, X2)) → PROPER(X2)
PROPER(len(X)) → PROPER(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(50) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(fst(X1, X2)) → PROPER(X1)
PROPER(fst(X1, X2)) → PROPER(X2)
PROPER(add(X1, X2)) → PROPER(X1)
PROPER(add(X1, X2)) → PROPER(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
fst(x1, x2)  =  fst(x1, x2)
from(x1)  =  x1
add(x1, x2)  =  add(x1, x2)
len(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
add2: [1,2]
cons2: [1,2]
PROPER1: [1]
fst2: [1,2]

The following usable rules [FROCOS05] were oriented: none

(51) Obligation:

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

PROPER(s(X)) → PROPER(X)
PROPER(from(X)) → PROPER(X)
PROPER(len(X)) → PROPER(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(52) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(len(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
s(x1)  =  x1
from(x1)  =  x1
len(x1)  =  len(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
PROPER1: [1]
len1: [1]

The following usable rules [FROCOS05] were oriented: none

(53) Obligation:

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

PROPER(s(X)) → PROPER(X)
PROPER(from(X)) → PROPER(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


PROPER(s(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
s(x1)  =  s(x1)
from(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
s1 > PROPER1

Status:
PROPER1: [1]
s1: [1]

The following usable rules [FROCOS05] were oriented: none

(55) Obligation:

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

PROPER(from(X)) → PROPER(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
from1: [1]

The following usable rules [FROCOS05] were oriented: none

(57) 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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(58) PisEmptyProof (EQUIVALENT transformation)

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

(59) TRUE

(60) Obligation:

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

ACTIVE(fst(X1, X2)) → ACTIVE(X1)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(fst(X1, X2)) → ACTIVE(X2)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(add(X1, X2)) → ACTIVE(X1)
ACTIVE(add(X1, X2)) → ACTIVE(X2)
ACTIVE(len(X)) → ACTIVE(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(61) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(fst(X1, X2)) → ACTIVE(X1)
ACTIVE(fst(X1, X2)) → ACTIVE(X2)
ACTIVE(add(X1, X2)) → ACTIVE(X1)
ACTIVE(add(X1, X2)) → ACTIVE(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE(x1)
fst(x1, x2)  =  fst(x1, x2)
cons(x1, x2)  =  x1
from(x1)  =  x1
add(x1, x2)  =  add(x1, x2)
len(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
add2: [1,2]
fst2: [1,2]
ACTIVE1: [1]

The following usable rules [FROCOS05] were oriented: none

(62) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(len(X)) → ACTIVE(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(63) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
ACTIVE1: [1]
len1: [1]

The following usable rules [FROCOS05] were oriented: none

(64) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(65) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(cons(X1, X2)) → ACTIVE(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE(x1)
cons(x1, x2)  =  cons(x1, x2)
from(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
cons2: [1,2]
ACTIVE1: [1]

The following usable rules [FROCOS05] were oriented: none

(66) Obligation:

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

ACTIVE(from(X)) → ACTIVE(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(67) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
from1: [1]

The following usable rules [FROCOS05] were oriented: none

(68) 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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(69) PisEmptyProof (EQUIVALENT transformation)

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

(70) TRUE

(71) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(72) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Precedence:
add2 > mark1
add2 > s
len1 > mark1
len1 > s
len1 > 0
fst2 > mark1
fst2 > nil
from1 > mark1
from1 > s

Status:
from1: [1]
add2: [1,2]
fst2: [1,2]
mark1: [1]
s: []
0: []
nil: []
len1: [1]

The following usable rules [FROCOS05] were oriented:

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

(73) Obligation:

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

TOP(ok(X)) → TOP(active(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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(74) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Precedence:
TOP1 > mark
len1 > ok1 > mark
s > cons2 > ok1 > mark
nil > mark
0 > mark

Status:
cons2: [2,1]
mark: []
ok1: [1]
TOP1: [1]
s: []
0: []
nil: []
len1: [1]

The following usable rules [FROCOS05] were oriented:

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

(75) 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)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(fst(X1, X2)) → fst(active(X1), X2)
active(fst(X1, X2)) → fst(X1, active(X2))
active(from(X)) → from(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(len(X)) → len(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
fst(mark(X1), X2) → mark(fst(X1, X2))
fst(X1, mark(X2)) → mark(fst(X1, X2))
from(mark(X)) → mark(from(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
len(mark(X)) → mark(len(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(fst(X1, X2)) → fst(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(len(X)) → len(proper(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
fst(ok(X1), ok(X2)) → ok(fst(X1, X2))
from(ok(X)) → ok(from(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
len(ok(X)) → ok(len(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(76) PisEmptyProof (EQUIVALENT transformation)

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

(77) TRUE