Termination w.r.t. Q proof of TRS/TRCSRExIntrod_GM04

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(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

TOP(mark(X)) → PROPER(X)
ACTIVE(zeros) → CONS(0, zeros)
ACTIVE(adx(cons(X, Y))) → CONS(X, adx(Y))
ACTIVE(tl(X)) → TL(active(X))
ACTIVE(incr(cons(X, Y))) → INCR(Y)
ACTIVE(tl(X)) → ACTIVE(X)
PROPER(adx(X)) → PROPER(X)
INCR(mark(X)) → INCR(X)
TL(ok(X)) → TL(X)
PROPER(adx(X)) → ADX(proper(X))
ACTIVE(adx(cons(X, Y))) → ADX(Y)
PROPER(incr(X)) → INCR(proper(X))
PROPER(incr(X)) → PROPER(X)
ACTIVE(incr(cons(X, Y))) → S(X)
TOP(mark(X)) → TOP(proper(X))
PROPER(s(X)) → S(proper(X))
S(ok(X)) → S(X)
ACTIVE(hd(X)) → ACTIVE(X)
ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(nats) → ADX(zeros)
PROPER(hd(X)) → HD(proper(X))
INCR(ok(X)) → INCR(X)
PROPER(hd(X)) → PROPER(X)
ADX(mark(X)) → ADX(X)
PROPER(cons(X1, X2)) → PROPER(X1)
ACTIVE(adx(cons(X, Y))) → INCR(cons(X, adx(Y)))
ACTIVE(incr(X)) → INCR(active(X))
ACTIVE(incr(cons(X, Y))) → CONS(s(X), incr(Y))
TOP(ok(X)) → ACTIVE(X)
ACTIVE(adx(X)) → ACTIVE(X)
PROPER(tl(X)) → TL(proper(X))
HD(mark(X)) → HD(X)
ADX(ok(X)) → ADX(X)
ACTIVE(hd(X)) → HD(active(X))
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(s(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
HD(ok(X)) → HD(X)
PROPER(cons(X1, X2)) → PROPER(X2)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(adx(X)) → ADX(active(X))
TL(mark(X)) → TL(X)
PROPER(tl(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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

      ↳ EdgeDeletionProof

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

TOP(mark(X)) → PROPER(X)
ACTIVE(zeros) → CONS(0, zeros)
ACTIVE(adx(cons(X, Y))) → CONS(X, adx(Y))
ACTIVE(tl(X)) → TL(active(X))
ACTIVE(incr(cons(X, Y))) → INCR(Y)
ACTIVE(tl(X)) → ACTIVE(X)
PROPER(adx(X)) → PROPER(X)
INCR(mark(X)) → INCR(X)
TL(ok(X)) → TL(X)
PROPER(adx(X)) → ADX(proper(X))
ACTIVE(adx(cons(X, Y))) → ADX(Y)
PROPER(incr(X)) → INCR(proper(X))
PROPER(incr(X)) → PROPER(X)
ACTIVE(incr(cons(X, Y))) → S(X)
TOP(mark(X)) → TOP(proper(X))
PROPER(s(X)) → S(proper(X))
S(ok(X)) → S(X)
ACTIVE(hd(X)) → ACTIVE(X)
ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(nats) → ADX(zeros)
PROPER(hd(X)) → HD(proper(X))
INCR(ok(X)) → INCR(X)
PROPER(hd(X)) → PROPER(X)
ADX(mark(X)) → ADX(X)
PROPER(cons(X1, X2)) → PROPER(X1)
ACTIVE(adx(cons(X, Y))) → INCR(cons(X, adx(Y)))
ACTIVE(incr(X)) → INCR(active(X))
ACTIVE(incr(cons(X, Y))) → CONS(s(X), incr(Y))
TOP(ok(X)) → ACTIVE(X)
ACTIVE(adx(X)) → ACTIVE(X)
PROPER(tl(X)) → TL(proper(X))
HD(mark(X)) → HD(X)
ADX(ok(X)) → ADX(X)
ACTIVE(hd(X)) → HD(active(X))
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(s(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
HD(ok(X)) → HD(X)
PROPER(cons(X1, X2)) → PROPER(X2)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(adx(X)) → ADX(active(X))
TL(mark(X)) → TL(X)
PROPER(tl(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

ACTIVE(zeros) → CONS(0, zeros)
TOP(mark(X)) → PROPER(X)
ACTIVE(adx(cons(X, Y))) → CONS(X, adx(Y))
ACTIVE(tl(X)) → TL(active(X))
ACTIVE(incr(cons(X, Y))) → INCR(Y)
PROPER(adx(X)) → PROPER(X)
ACTIVE(tl(X)) → ACTIVE(X)
INCR(mark(X)) → INCR(X)
PROPER(adx(X)) → ADX(proper(X))
TL(ok(X)) → TL(X)
ACTIVE(adx(cons(X, Y))) → ADX(Y)
PROPER(incr(X)) → INCR(proper(X))
ACTIVE(incr(cons(X, Y))) → S(X)
PROPER(incr(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
TOP(mark(X)) → TOP(proper(X))
ACTIVE(hd(X)) → ACTIVE(X)
S(ok(X)) → S(X)
ACTIVE(nats) → ADX(zeros)
ACTIVE(incr(X)) → ACTIVE(X)
PROPER(hd(X)) → HD(proper(X))
INCR(ok(X)) → INCR(X)
ADX(mark(X)) → ADX(X)
PROPER(hd(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
ACTIVE(adx(cons(X, Y))) → INCR(cons(X, adx(Y)))
ACTIVE(incr(X)) → INCR(active(X))
ACTIVE(incr(cons(X, Y))) → CONS(s(X), incr(Y))
TOP(ok(X)) → ACTIVE(X)
ACTIVE(adx(X)) → ACTIVE(X)
PROPER(tl(X)) → TL(proper(X))
HD(mark(X)) → HD(X)
ACTIVE(hd(X)) → HD(active(X))
ADX(ok(X)) → ADX(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(s(X)) → PROPER(X)
HD(ok(X)) → HD(X)
TOP(ok(X)) → TOP(active(X))
PROPER(cons(X1, X2)) → PROPER(X2)
ACTIVE(adx(X)) → ADX(active(X))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
TL(mark(X)) → TL(X)
PROPER(tl(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 [13,14,18] contains 9 SCCs with 20 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

S(ok(X)) → S(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

S(ok(X)) → S(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
S(x1) = x1
ok(x1) = ok(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = x2
ok(x1) = ok(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

TL(ok(X)) → TL(X)
TL(mark(X)) → TL(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

TL(mark(X)) → TL(X)

The remaining pairs can at least be oriented weakly.

TL(ok(X)) → TL(X)

Used ordering: Combined order from the following AFS and order.
TL(x1) = x1
ok(x1) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

TL(ok(X)) → TL(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

TL(ok(X)) → TL(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
TL(x1) = x1
ok(x1) = ok(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

HD(mark(X)) → HD(X)
HD(ok(X)) → HD(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

HD(ok(X)) → HD(X)

The remaining pairs can at least be oriented weakly.

HD(mark(X)) → HD(X)

Used ordering: Combined order from the following AFS and order.
HD(x1) = x1
mark(x1) = x1
ok(x1) = ok(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

HD(mark(X)) → HD(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

HD(mark(X)) → HD(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
HD(x1) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

INCR(ok(X)) → INCR(X)
INCR(mark(X)) → INCR(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

INCR(mark(X)) → INCR(X)

The remaining pairs can at least be oriented weakly.

INCR(ok(X)) → INCR(X)

Used ordering: Combined order from the following AFS and order.
INCR(x1) = x1
ok(x1) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

INCR(ok(X)) → INCR(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

INCR(ok(X)) → INCR(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
INCR(x1) = x1
ok(x1) = ok(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

ADX(ok(X)) → ADX(X)
ADX(mark(X)) → ADX(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ADX(mark(X)) → ADX(X)

The remaining pairs can at least be oriented weakly.

ADX(ok(X)) → ADX(X)

Used ordering: Combined order from the following AFS and order.
ADX(x1) = x1
ok(x1) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

ADX(ok(X)) → ADX(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ADX(ok(X)) → ADX(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
ADX(x1) = x1
ok(x1) = ok(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

PROPER(hd(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(incr(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(adx(X)) → PROPER(X)
PROPER(tl(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)

The remaining pairs can at least be oriented weakly.

PROPER(hd(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(incr(X)) → PROPER(X)
PROPER(adx(X)) → PROPER(X)
PROPER(tl(X)) → PROPER(X)

Used ordering: Combined order from the following AFS and order.
PROPER(x1) = x1
hd(x1) = x1
s(x1) = x1
cons(x1, x2) = cons(x1, x2)
incr(x1) = x1
adx(x1) = x1
tl(x1) = x1

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

PROPER(s(X)) → PROPER(X)
PROPER(hd(X)) → PROPER(X)
PROPER(incr(X)) → PROPER(X)
PROPER(adx(X)) → PROPER(X)
PROPER(tl(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

PROPER(tl(X)) → PROPER(X)

The remaining pairs can at least be oriented weakly.

PROPER(s(X)) → PROPER(X)
PROPER(hd(X)) → PROPER(X)
PROPER(incr(X)) → PROPER(X)
PROPER(adx(X)) → PROPER(X)

Used ordering: Combined order from the following AFS and order.
PROPER(x1) = x1
s(x1) = x1
hd(x1) = x1
incr(x1) = x1
adx(x1) = x1
tl(x1) = tl(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

PROPER(hd(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(incr(X)) → PROPER(X)
PROPER(adx(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

PROPER(adx(X)) → PROPER(X)

The remaining pairs can at least be oriented weakly.

PROPER(hd(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(incr(X)) → PROPER(X)

Used ordering: Combined order from the following AFS and order.
PROPER(x1) = x1
hd(x1) = x1
s(x1) = x1
incr(x1) = x1
adx(x1) = adx(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

PROPER(s(X)) → PROPER(X)
PROPER(hd(X)) → PROPER(X)
PROPER(incr(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

PROPER(incr(X)) → PROPER(X)

The remaining pairs can at least be oriented weakly.

PROPER(s(X)) → PROPER(X)
PROPER(hd(X)) → PROPER(X)

Used ordering: Combined order from the following AFS and order.
PROPER(x1) = x1
s(x1) = x1
hd(x1) = x1
incr(x1) = incr(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

PROPER(hd(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

PROPER(s(X)) → PROPER(X)

The remaining pairs can at least be oriented weakly.

PROPER(hd(X)) → PROPER(X)

Used ordering: Combined order from the following AFS and order.
PROPER(x1) = x1
hd(x1) = x1
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

PROPER(hd(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

PROPER(hd(X)) → PROPER(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PROPER(x1) = x1
hd(x1) = hd(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

ACTIVE(hd(X)) → ACTIVE(X)
ACTIVE(tl(X)) → ACTIVE(X)
ACTIVE(adx(X)) → ACTIVE(X)
ACTIVE(incr(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ACTIVE(incr(X)) → ACTIVE(X)

The remaining pairs can at least be oriented weakly.

ACTIVE(hd(X)) → ACTIVE(X)
ACTIVE(tl(X)) → ACTIVE(X)
ACTIVE(adx(X)) → ACTIVE(X)

Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
hd(x1) = x1
tl(x1) = x1
adx(x1) = x1
incr(x1) = incr(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

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

ACTIVE(hd(X)) → ACTIVE(X)
ACTIVE(tl(X)) → ACTIVE(X)
ACTIVE(adx(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ACTIVE(adx(X)) → ACTIVE(X)

The remaining pairs can at least be oriented weakly.

ACTIVE(hd(X)) → ACTIVE(X)
ACTIVE(tl(X)) → ACTIVE(X)

Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
hd(x1) = x1
tl(x1) = x1
adx(x1) = adx(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

              ↳ QDP

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

ACTIVE(hd(X)) → ACTIVE(X)
ACTIVE(tl(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ACTIVE(tl(X)) → ACTIVE(X)

The remaining pairs can at least be oriented weakly.

ACTIVE(hd(X)) → ACTIVE(X)

Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
hd(x1) = x1
tl(x1) = tl(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

              ↳ QDP

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

ACTIVE(hd(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ACTIVE(hd(X)) → ACTIVE(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
hd(x1) = hd(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ PisEmptyProof

              ↳ QDP

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

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

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
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.