Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE

(0) Obligation:

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

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
activate(n__fib1(X1, X2)) → fib1(X1, X2)
activate(X) → X

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

FIB(N) → SEL(N, fib1(s(0), s(0)))
FIB(N) → FIB1(s(0), s(0))
FIB1(X, Y) → ADD(X, Y)
ADD(s(X), Y) → ADD(X, Y)
SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
SEL(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__fib1(X1, X2)) → FIB1(X1, X2)

The TRS R consists of the following rules:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
activate(n__fib1(X1, X2)) → fib1(X1, X2)
activate(X) → X

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
activate(n__fib1(X1, X2)) → fib1(X1, X2)
activate(X) → X

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive path order with status [RPO].
Precedence:
s1 > ADD1

Status:
s1: multiset
ADD1: multiset

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
activate(n__fib1(X1, X2)) → fib1(X1, X2)
activate(X) → X

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))

The TRS R consists of the following rules:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
activate(n__fib1(X1, X2)) → fib1(X1, X2)
activate(X) → X

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  SEL(x1)
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x2)
activate(x1)  =  activate
n__fib1(x1, x2)  =  n__fib1(x1, x2)
fib1(x1, x2)  =  fib1(x1)
add(x1, x2)  =  add
0  =  0

Recursive path order with status [RPO].
Precedence:
SEL1 > s1
activate > s1
fib11 > cons1 > s1
fib11 > nfib12 > s1
add > s1
0 > s1

Status:
cons1: multiset
add: multiset
fib11: multiset
nfib12: multiset
activate: multiset
s1: multiset
SEL1: multiset
0: multiset

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
activate(n__fib1(X1, X2)) → fib1(X1, X2)
activate(X) → X

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE