Termination w.r.t. Q proof of /tmp/tmpsENSQd/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

(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, 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)
active(x1) = x1

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

[ADD₂, mark₁]

Status:

mark₁: [1]
ADD₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:

[ADD₂, active₁]

Status:

active₁: [1]
ADD₂: [2,1]

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))
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) 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(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.

(13) 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)
CONS(mark(X1), X2) → CONS(X1, X2)

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

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

[CONS₂, mark₁]

Status:

CONS₂: [2,1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

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

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:

[CONS₂, active₁]

Status:

active₁: [1]
CONS₂: [2,1]

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))
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) 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(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.

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

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

[S₁, active₁]

Status:

active₁: [1]
S₁: [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))
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.

(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))
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) 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(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.

(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)
active(x1) = x1

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

[FIB1₂, mark₁]

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(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.

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FIB1(active(X1), X2) → FIB1(X1, X2)
FIB1(X1, active(X2)) → FIB1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:

[FIB1₂, active₁]

Status:

active₁: [1]
FIB1₂: [2,1]

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))
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) 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(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.

(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)
active(x1) = x1

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

[SEL₂, mark₁]

Status:

mark₁: [1]
SEL₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

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

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.

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL(active(X1), X2) → SEL(X1, X2)
SEL(X1, active(X2)) → SEL(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:

[SEL₂, active₁]

Status:

active₁: [1]
SEL₂: [2,1]

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))
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.

(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(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.

(41) 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) = FIB(x1)
active(x1) = active(x1)
mark(x1) = x1

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

[FIB₁, active₁]

Status:

active₁: [1]
FIB₁: [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))
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.

(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))
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.

(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:

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.

(48) 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) = x1
sel(x1, x2) = sel
fib1(x1, x2) = fib1
s(x1) = s
0 = 0
cons(x1, x2) = cons
add(x1, x2) = add
active(x1) = active

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

[MARK, fib, sel, fib1, s, add] > [0, active] > cons

Status:

active: []
MARK: []
sel: []
cons: []
add: []
fib: []
fib1: []
s: []
0: []

The following usable rules [FROCOS05] were oriented:

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

(49) 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.

(50) 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
add(x1, x2) = add
active(x1) = active

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

cons > [MARK, fib, sel, fib1, add] > [mark₁, s, 0] > active

Status:

active: []
MARK: []
sel: []
cons: []
add: []
fib: []
fib1: []
mark₁: [1]
s: []
0: []

The following usable rules [FROCOS05] were oriented:

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

(51) 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.