Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 QDPOrderProof (⇔)
    ↳11 QDP
    ↳12 PisEmptyProof (⇔)
    ↳13 TRUE
  ↳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 QDPOrderProof (⇔)
    ↳43 QDP
    ↳44 PisEmptyProof (⇔)
    ↳45 TRUE
  ↳46 QDP
    ↳47 QDPOrderProof (⇔)
    ↳48 QDP
    ↳49 QDPOrderProof (⇔)
    ↳50 QDP
    ↳51 PisEmptyProof (⇔)
    ↳52 TRUE
  ↳53 QDP
    ↳54 QDPOrderProof (⇔)
    ↳55 QDP
    ↳56 QDPOrderProof (⇔)
    ↳57 QDP
    ↳58 QDPOrderProof (⇔)
    ↳59 QDP
    ↳60 PisEmptyProof (⇔)
    ↳61 TRUE
  ↳62 QDP
    ↳63 QDPOrderProof (⇔)
    ↳64 QDP
    ↳65 QDPOrderProof (⇔)
    ↳66 QDP
    ↳67 QDPOrderProof (⇔)
    ↳68 QDP
    ↳69 QDPOrderProof (⇔)
    ↳70 QDP
    ↳71 PisEmptyProof (⇔)
    ↳72 TRUE
  ↳73 QDP

(0) Obligation:

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

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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(fib(N)) → SEL(N, fib1(s(0), s(0)))
ACTIVE(fib(N)) → FIB1(s(0), s(0))
ACTIVE(fib(N)) → S(0)
ACTIVE(fib1(X, Y)) → CONS(X, fib1(Y, add(X, Y)))
ACTIVE(fib1(X, Y)) → FIB1(Y, add(X, Y))
ACTIVE(fib1(X, Y)) → ADD(X, Y)
ACTIVE(add(s(X), Y)) → S(add(X, Y))
ACTIVE(add(s(X), Y)) → ADD(X, Y)
ACTIVE(sel(s(N), cons(X, XS))) → SEL(N, XS)
ACTIVE(fib(X)) → FIB(active(X))
ACTIVE(fib(X)) → ACTIVE(X)
ACTIVE(sel(X1, X2)) → SEL(active(X1), X2)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → SEL(X1, active(X2))
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(fib1(X1, X2)) → FIB1(active(X1), X2)
ACTIVE(fib1(X1, X2)) → ACTIVE(X1)
ACTIVE(fib1(X1, X2)) → FIB1(X1, active(X2))
ACTIVE(fib1(X1, X2)) → ACTIVE(X2)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
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)
FIB(mark(X)) → FIB(X)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)
FIB1(mark(X1), X2) → FIB1(X1, X2)
FIB1(X1, mark(X2)) → FIB1(X1, X2)
S(mark(X)) → S(X)
CONS(mark(X1), X2) → CONS(X1, X2)
ADD(mark(X1), X2) → ADD(X1, X2)
ADD(X1, mark(X2)) → ADD(X1, X2)
PROPER(fib(X)) → FIB(proper(X))
PROPER(fib(X)) → PROPER(X)
PROPER(sel(X1, X2)) → SEL(proper(X1), proper(X2))
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(fib1(X1, X2)) → FIB1(proper(X1), proper(X2))
PROPER(fib1(X1, X2)) → PROPER(X1)
PROPER(fib1(X1, X2)) → PROPER(X2)
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(add(X1, X2)) → ADD(proper(X1), proper(X2))
PROPER(add(X1, X2)) → PROPER(X1)
PROPER(add(X1, X2)) → PROPER(X2)
FIB(ok(X)) → FIB(X)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
FIB1(ok(X1), ok(X2)) → FIB1(X1, X2)
S(ok(X)) → S(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ADD(ok(X1), ok(X2)) → ADD(X1, X2)
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(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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 26 less nodes.

(4) Complex Obligation (AND)

(5) 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(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.


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

Lexicographic Path Order [LPO].
Precedence:
mark1 > ADD1

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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) 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)  =  x2
mark(x1)  =  mark
ok(x1)  =  ok(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

ADD(mark(X1), X2) → ADD(X1, X2)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.

(10) 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)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

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

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) 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(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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) 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(x2)
ok(x1)  =  ok(x1)
mark(x1)  =  mark(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:

CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.

(17) 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)
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(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.

(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:

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.


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

Lexicographic Path Order [LPO].
Precedence:
mark1 > S1

The following usable rules [FROCOS05] were oriented: none

(23) 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(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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) 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 [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(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.

(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:

FIB1(X1, mark(X2)) → FIB1(X1, X2)
FIB1(mark(X1), X2) → FIB1(X1, X2)
FIB1(ok(X1), ok(X2)) → FIB1(X1, X2)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.


FIB1(X1, mark(X2)) → FIB1(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIB1(x1, x2)  =  FIB1(x2)
mark(x1)  =  mark(x1)
ok(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
mark1 > FIB11

The following usable rules [FROCOS05] were oriented: none

(30) Obligation:

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

FIB1(mark(X1), X2) → FIB1(X1, X2)
FIB1(ok(X1), ok(X2)) → FIB1(X1, X2)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FIB1(ok(X1), ok(X2)) → FIB1(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIB1(x1, x2)  =  x2
mark(x1)  =  mark
ok(x1)  =  ok(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:

FIB1(mark(X1), X2) → FIB1(X1, X2)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FIB1(mark(X1), X2) → FIB1(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIB1(x1, x2)  =  FIB1(x1)
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(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.

(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:

SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.


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

Lexicographic Path Order [LPO].
Precedence:
mark1 > SEL1

The following usable rules [FROCOS05] were oriented: none

(39) Obligation:

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

SEL(mark(X1), X2) → SEL(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(41) Obligation:

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

SEL(mark(X1), X2) → SEL(X1, X2)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.

(42) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(43) Obligation:

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

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.

(44) PisEmptyProof (EQUIVALENT transformation)

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

(45) TRUE

(46) Obligation:

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

FIB(ok(X)) → FIB(X)
FIB(mark(X)) → FIB(X)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
mark1 > FIB1

The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

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

FIB(ok(X)) → FIB(X)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FIB(ok(X)) → FIB(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIB(x1)  =  x1
ok(x1)  =  ok(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(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.

(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:

PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(fib(X)) → PROPER(X)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(fib1(X1, X2)) → PROPER(X1)
PROPER(fib1(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(add(X1, X2)) → PROPER(X1)
PROPER(add(X1, X2)) → PROPER(X2)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(fib1(X1, X2)) → PROPER(X1)
PROPER(fib1(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(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)  =  x1
sel(x1, x2)  =  sel(x1, x2)
fib(x1)  =  x1
fib1(x1, x2)  =  fib1(x1, x2)
s(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
add(x1, x2)  =  add(x1, x2)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(55) Obligation:

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

PROPER(fib(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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(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)
fib(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
s1 > PROPER1

The following usable rules [FROCOS05] were oriented: none

(57) Obligation:

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

PROPER(fib(X)) → PROPER(X)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(fib(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
fib(x1)  =  fib(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

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(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.

(60) PisEmptyProof (EQUIVALENT transformation)

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

(61) TRUE

(62) Obligation:

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

ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(fib(X)) → ACTIVE(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(fib1(X1, X2)) → ACTIVE(X1)
ACTIVE(fib1(X1, X2)) → ACTIVE(X2)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(add(X1, X2)) → ACTIVE(X1)
ACTIVE(add(X1, X2)) → ACTIVE(X2)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(fib1(X1, X2)) → ACTIVE(X1)
ACTIVE(fib1(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)
sel(x1, x2)  =  sel(x1, x2)
fib(x1)  =  x1
fib1(x1, x2)  =  fib1(x1, x2)
s(x1)  =  x1
cons(x1, x2)  =  x1
add(x1, x2)  =  add(x1, x2)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(64) Obligation:

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

ACTIVE(fib(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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(fib(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
fib(x1)  =  fib(x1)
s(x1)  =  x1
cons(x1, x2)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(66) Obligation:

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

ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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(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)
s(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(68) Obligation:

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

ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(s(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
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(70) Obligation:

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

active(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.

(71) PisEmptyProof (EQUIVALENT transformation)

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

(72) TRUE

(73) 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(fib(N)) → mark(sel(N, fib1(s(0), s(0))))
active(fib1(X, Y)) → mark(cons(X, fib1(Y, add(X, Y))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(fib(X)) → fib(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(fib1(X1, X2)) → fib1(active(X1), X2)
active(fib1(X1, X2)) → fib1(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
fib(mark(X)) → mark(fib(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
fib1(mark(X1), X2) → mark(fib1(X1, X2))
fib1(X1, mark(X2)) → mark(fib1(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
proper(fib(X)) → fib(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(fib1(X1, X2)) → fib1(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
fib(ok(X)) → ok(fib(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
fib1(ok(X1), ok(X2)) → ok(fib1(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
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.