Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 9 SCCs with 26 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesReductionPairsProof

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

TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(TOP(x1)) = x1   
POL(active(x1)) = 2·x1   
POL(add(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(fib(x1)) = 2·x1   
POL(fib1(x1, x2)) = 2·x1 + 2·x2   
POL(mark(x1)) = x1   
POL(ok(x1)) = 2·x1   
POL(proper(x1)) = x1   
POL(s(x1)) = 2·x1   
POL(sel(x1, x2)) = x1 + 2·x2   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesReductionPairsProof
QDP
                ↳ Narrowing

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

TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule TOP(mark(X)) → TOP(proper(X)) at position [0] we obtained the following new rules:

TOP(mark(fib(x0))) → TOP(fib(proper(x0)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(fib1(x0, x1))) → TOP(fib1(proper(x0), proper(x1)))
TOP(mark(add(x0, x1))) → TOP(add(proper(x0), proper(x1)))
TOP(mark(0)) → TOP(ok(0))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesReductionPairsProof
              ↳ QDP
                ↳ Narrowing
QDP
                    ↳ Narrowing

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

TOP(mark(fib(x0))) → TOP(fib(proper(x0)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(fib1(x0, x1))) → TOP(fib1(proper(x0), proper(x1)))
TOP(mark(add(x0, x1))) → TOP(add(proper(x0), proper(x1)))
TOP(mark(0)) → TOP(ok(0))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule TOP(ok(X)) → TOP(active(X)) at position [0] we obtained the following new rules:

TOP(ok(add(0, x0))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(ok(fib1(x0, x1))) → TOP(mark(cons(x0, fib1(x1, add(x0, x1)))))
TOP(ok(add(s(x0), x1))) → TOP(mark(s(add(x0, x1))))
TOP(ok(fib(x0))) → TOP(mark(sel(x0, fib1(s(0), s(0)))))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(fib1(x0, x1))) → TOP(fib1(active(x0), x1))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(ok(fib1(x0, x1))) → TOP(fib1(x0, active(x1)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(add(x0, x1))) → TOP(add(x0, active(x1)))
TOP(ok(fib(x0))) → TOP(fib(active(x0)))
TOP(ok(add(x0, x1))) → TOP(add(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesReductionPairsProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ Narrowing
QDP
                        ↳ DependencyGraphProof

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

TOP(mark(fib(x0))) → TOP(fib(proper(x0)))
TOP(ok(add(0, x0))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(fib1(x0, x1))) → TOP(mark(cons(x0, fib1(x1, add(x0, x1)))))
TOP(ok(add(s(x0), x1))) → TOP(mark(s(add(x0, x1))))
TOP(mark(0)) → TOP(ok(0))
TOP(ok(fib(x0))) → TOP(mark(sel(x0, fib1(s(0), s(0)))))
TOP(ok(fib1(x0, x1))) → TOP(fib1(active(x0), x1))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(fib1(x0, x1))) → TOP(fib1(x0, active(x1)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(add(x0, x1))) → TOP(add(x0, active(x1)))
TOP(ok(fib(x0))) → TOP(fib(active(x0)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(add(x0, x1))) → TOP(add(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(fib1(x0, x1))) → TOP(fib1(proper(x0), proper(x1)))
TOP(mark(add(x0, x1))) → TOP(add(proper(x0), proper(x1)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesReductionPairsProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ Narrowing
                      ↳ QDP
                        ↳ DependencyGraphProof
QDP

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

TOP(mark(fib(x0))) → TOP(fib(proper(x0)))
TOP(ok(add(0, x0))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(fib1(x0, x1))) → TOP(mark(cons(x0, fib1(x1, add(x0, x1)))))
TOP(ok(add(s(x0), x1))) → TOP(mark(s(add(x0, x1))))
TOP(ok(fib(x0))) → TOP(mark(sel(x0, fib1(s(0), s(0)))))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(fib1(x0, x1))) → TOP(fib1(active(x0), x1))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(fib1(x0, x1))) → TOP(fib1(x0, active(x1)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(add(x0, x1))) → TOP(add(x0, active(x1)))
TOP(ok(fib(x0))) → TOP(fib(active(x0)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(add(x0, x1))) → TOP(add(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(fib1(x0, x1))) → TOP(fib1(proper(x0), proper(x1)))
TOP(mark(add(x0, x1))) → TOP(add(proper(x0), proper(x1)))

The TRS R consists of the following rules:

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

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