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 QDPOrderProof (⇔)
    ↳20 QDP
    ↳21 PisEmptyProof (⇔)
    ↳22 TRUE
  ↳23 QDP
    ↳24 QDPOrderProof (⇔)
    ↳25 QDP
    ↳26 QDPOrderProof (⇔)
    ↳27 QDP
    ↳28 PisEmptyProof (⇔)
    ↳29 TRUE
  ↳30 QDP
    ↳31 QDPOrderProof (⇔)
    ↳32 QDP
    ↳33 QDPOrderProof (⇔)
    ↳34 QDP
    ↳35 QDPOrderProof (⇔)
    ↳36 QDP
    ↳37 PisEmptyProof (⇔)
    ↳38 TRUE
  ↳39 QDP
    ↳40 QDPOrderProof (⇔)
    ↳41 QDP
    ↳42 QDPOrderProof (⇔)
    ↳43 QDP
    ↳44 QDPOrderProof (⇔)
    ↳45 QDP
    ↳46 PisEmptyProof (⇔)
    ↳47 TRUE
  ↳48 QDP
    ↳49 QDPOrderProof (⇔)
    ↳50 QDP
    ↳51 QDPOrderProof (⇔)
    ↳52 QDP
    ↳53 PisEmptyProof (⇔)
    ↳54 TRUE
  ↳55 QDP
    ↳56 QDPOrderProof (⇔)
    ↳57 QDP
    ↳58 QDPOrderProof (⇔)
    ↳59 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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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)) → MARK(sel(N, fib1(s(0), s(0))))
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)) → MARK(cons(X, fib1(Y, add(X, Y))))
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(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
ACTIVE(add(s(X), Y)) → S(add(X, Y))
ACTIVE(add(s(X), Y)) → ADD(X, Y)
ACTIVE(sel(0, cons(X, XS))) → MARK(X)
ACTIVE(sel(s(N), cons(X, XS))) → MARK(sel(N, XS))
ACTIVE(sel(s(N), cons(X, XS))) → SEL(N, XS)
MARK(fib(X)) → ACTIVE(fib(mark(X)))
MARK(fib(X)) → FIB(mark(X))
MARK(fib(X)) → MARK(X)
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
MARK(sel(X1, X2)) → SEL(mark(X1), mark(X2))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)
MARK(fib1(X1, X2)) → ACTIVE(fib1(mark(X1), mark(X2)))
MARK(fib1(X1, X2)) → FIB1(mark(X1), mark(X2))
MARK(fib1(X1, X2)) → MARK(X1)
MARK(fib1(X1, X2)) → MARK(X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(0) → ACTIVE(0)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(add(X1, X2)) → ADD(mark(X1), mark(X2))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
FIB(mark(X)) → FIB(X)
FIB(active(X)) → FIB(X)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(active(X1), X2) → SEL(X1, X2)
SEL(X1, active(X2)) → SEL(X1, X2)
FIB1(mark(X1), X2) → FIB1(X1, X2)
FIB1(X1, mark(X2)) → FIB1(X1, X2)
FIB1(active(X1), X2) → FIB1(X1, X2)
FIB1(X1, active(X2)) → FIB1(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(X)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
ADD(mark(X1), X2) → ADD(X1, X2)
ADD(X1, mark(X2)) → ADD(X1, X2)
ADD(active(X1), X2) → ADD(X1, X2)
ADD(X1, active(X2)) → ADD(X1, X2)

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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 16 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(active(X1), X2) → ADD(X1, X2)
ADD(X1, active(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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

The TRS R consists of the following rules:

active(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

ADD(X1, mark(X2)) → ADD(X1, X2)

The TRS R consists of the following rules:

active(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

The TRS R consists of the following rules:

active(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

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

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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

CONS(X1, mark(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(20) 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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(25) 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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(27) 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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE

(30) 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(active(X1), X2) → FIB1(X1, X2)
FIB1(X1, active(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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(X1, active(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)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(32) Obligation:

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

FIB1(X1, mark(X2)) → FIB1(X1, X2)
FIB1(mark(X1), X2) → FIB1(X1, X2)
FIB1(active(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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)
FIB1(active(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)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(34) Obligation:

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

FIB1(X1, mark(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(35) 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)  =  x2
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(36) 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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(37) PisEmptyProof (EQUIVALENT transformation)

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

(38) TRUE

(39) 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(active(X1), X2) → SEL(X1, X2)
SEL(X1, active(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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(X1, active(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)  =  x1
active(x1)  =  active(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(X1, mark(X2)) → SEL(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(active(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(43) Obligation:

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

SEL(X1, mark(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(44) 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)  =  x2
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(45) 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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) Obligation:

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

FIB(active(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(50) 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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(51) 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 [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(52) 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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(53) PisEmptyProof (EQUIVALENT transformation)

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

(54) TRUE

(55) Obligation:

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

MARK(fib(X)) → ACTIVE(fib(mark(X)))
ACTIVE(fib(N)) → MARK(sel(N, fib1(s(0), s(0))))
MARK(fib(X)) → MARK(X)
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
ACTIVE(fib1(X, Y)) → MARK(cons(X, fib1(Y, add(X, Y))))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)
MARK(fib1(X1, X2)) → ACTIVE(fib1(mark(X1), mark(X2)))
ACTIVE(add(0, X)) → MARK(X)
MARK(fib1(X1, X2)) → MARK(X1)
MARK(fib1(X1, X2)) → MARK(X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(sel(0, cons(X, XS))) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(sel(s(N), cons(X, XS))) → MARK(sel(N, XS))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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.


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

Lexicographic Path Order [LPO].
Precedence:
[MARK, fib, mark, sel, fib1, s, add] > [0, cons, active1]


The following usable rules [FROCOS05] were oriented:

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

(57) Obligation:

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

MARK(fib(X)) → ACTIVE(fib(mark(X)))
ACTIVE(fib(N)) → MARK(sel(N, fib1(s(0), s(0))))
MARK(fib(X)) → MARK(X)
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
ACTIVE(fib1(X, Y)) → MARK(cons(X, fib1(Y, add(X, Y))))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)
MARK(fib1(X1, X2)) → ACTIVE(fib1(mark(X1), mark(X2)))
ACTIVE(add(0, X)) → MARK(X)
MARK(fib1(X1, X2)) → MARK(X1)
MARK(fib1(X1, X2)) → MARK(X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(X)) → MARK(X)
ACTIVE(sel(0, cons(X, XS))) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(sel(s(N), cons(X, XS))) → MARK(sel(N, XS))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(58) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(s(X)) → ACTIVE(s(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK
fib(x1)  =  fib
ACTIVE(x1)  =  x1
mark(x1)  =  mark(x1)
sel(x1, x2)  =  sel
fib1(x1, x2)  =  fib1
s(x1)  =  s
0  =  0
cons(x1, x2)  =  cons(x1, x2)
add(x1, x2)  =  add
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
[MARK, fib, mark1, sel, fib1, add] > [s, 0, cons2, active1]


The following usable rules [FROCOS05] were oriented:

fib(active(X)) → fib(X)
fib(mark(X)) → fib(X)
sel(X1, mark(X2)) → sel(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)

(59) Obligation:

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

MARK(fib(X)) → ACTIVE(fib(mark(X)))
ACTIVE(fib(N)) → MARK(sel(N, fib1(s(0), s(0))))
MARK(fib(X)) → MARK(X)
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
ACTIVE(fib1(X, Y)) → MARK(cons(X, fib1(Y, add(X, Y))))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)
MARK(fib1(X1, X2)) → ACTIVE(fib1(mark(X1), mark(X2)))
ACTIVE(add(0, X)) → MARK(X)
MARK(fib1(X1, X2)) → MARK(X1)
MARK(fib1(X1, X2)) → MARK(X2)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(X)) → MARK(X)
ACTIVE(sel(0, cons(X, XS))) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(sel(s(N), cons(X, XS))) → MARK(sel(N, XS))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(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))
mark(fib(X)) → active(fib(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(fib1(X1, X2)) → active(fib1(mark(X1), mark(X2)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
fib(mark(X)) → fib(X)
fib(active(X)) → fib(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
fib1(mark(X1), X2) → fib1(X1, X2)
fib1(X1, mark(X2)) → fib1(X1, X2)
fib1(active(X1), X2) → fib1(X1, X2)
fib1(X1, active(X2)) → fib1(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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