Termination w.r.t. Q proof of /tmp/tmpB5q3V6/Ex8_BLR02

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 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 PisEmptyProof (⇔)

    ↳18 TRUE

  ↳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 PisEmptyProof (⇔)

    ↳39 TRUE

  ↳40 QDP

    ↳41 QDPOrderProof (⇔)

    ↳42 QDP

    ↳43 QDPOrderProof (⇔)

    ↳44 QDP

    ↳45 PisEmptyProof (⇔)

    ↳46 TRUE

  ↳47 QDP

    ↳48 QDPOrderProof (⇔)

    ↳49 QDP

    ↳50 QDPOrderProof (⇔)

    ↳51 QDP

    ↳52 PisEmptyProof (⇔)

    ↳53 TRUE

  ↳54 QDP

    ↳55 QDPOrderProof (⇔)

    ↳56 QDP

    ↳57 QDPOrderProof (⇔)

    ↳58 QDP

    ↳59 QDPOrderProof (⇔)

    ↳60 QDP

    ↳61 QDPOrderProof (⇔)

    ↳62 QDP

    ↳63 PisEmptyProof (⇔)

    ↳64 TRUE

  ↳65 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)
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, x2)
mark(x1) = mark(x1)
ok(x1) = x1

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

trivial

Status:

ADD₂: [2,1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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) = ADD(x2)
ok(x1) = ok(x1)

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

trivial

Status:

ADD₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

active(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) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

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.

(13) 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, x2)
ok(x1) = ok(x1)
mark(x1) = x1

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

[CONS₂, ok₁]

Status:

CONS₂: [2,1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) 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.

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

[CONS₂, mark₁]

Status:

CONS₂: [2,1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(16) 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.

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) 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.

(20) 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) = S(x1)
ok(x1) = ok(x1)
mark(x1) = x1

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

[S₁, ok₁]

Status:

S₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

S(mark(X)) → S(X)

The TRS R consists of the following rules:

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

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

trivial

Status:

mark₁: [1]

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(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) 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:

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.

(27) 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)
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, x2)
mark(x1) = mark(x1)
ok(x1) = x1

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

trivial

Status:

FIB1₂: [2,1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

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(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) = FIB1(x2)
ok(x1) = ok(x1)

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

trivial

Status:

FIB1₁: [1]
ok₁: [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(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) 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:

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.

(34) 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)
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, x2)
mark(x1) = mark(x1)
ok(x1) = x1

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

trivial

Status:

SEL₂: [2,1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

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

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.

(36) 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) = SEL(x2)
ok(x1) = ok(x1)

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

trivial

Status:

SEL₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(37) 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.

(38) PisEmptyProof (EQUIVALENT transformation)

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

(39) TRUE

(40) 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.

(41) 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) = FIB(x1)
ok(x1) = ok(x1)
mark(x1) = x1

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

[FIB₁, ok₁]

Status:

FIB₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(42) Obligation:

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

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.

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

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

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(44) 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.

(45) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE

(47) 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.

(48) 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(fib(X)) → PROPER(X)
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) = PROPER(x1)
sel(x1, x2) = sel(x1, x2)
fib(x1) = fib(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 with status [LPO].
Quasi-Precedence:

trivial

Status:

PROPER₁: [1]
sel₂: [2,1]
fib₁: [1]
fib1₂: [2,1]
cons₂: [1,2]
add₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(49) Obligation:

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

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.

(50) 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) = x1
s(x1) = s(x1)

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

trivial

Status:

s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(51) 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.

(52) PisEmptyProof (EQUIVALENT transformation)

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

(53) TRUE

(54) 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.

(55) 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 with status [LPO].
Quasi-Precedence:

trivial

Status:

ACTIVE₁: [1]
sel₂: [1,2]
fib1₂: [1,2]
add₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(56) 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.

(57) 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)
fib(x1) = x1
s(x1) = x1
cons(x1, x2) = cons(x1, x2)

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

[ACTIVE₁, cons₂]

Status:

ACTIVE₁: [1]
cons₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(58) Obligation:

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

ACTIVE(fib(X)) → ACTIVE(X)
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.

(59) 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) = ACTIVE(x1)
fib(x1) = fib(x1)
s(x1) = x1

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

[ACTIVE₁, fib₁]

Status:

ACTIVE₁: [1]
fib₁: [1]

The following usable rules [FROCOS05] were oriented: none

(60) 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.

(61) 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 with status [LPO].
Quasi-Precedence:

trivial

Status:

s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(62) 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.

(63) PisEmptyProof (EQUIVALENT transformation)

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

(64) TRUE

(65) 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.