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:

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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:

TOP₁(mark₁(X)) -> PROPER₁(X)
ACTIVE₁(zeros) -> CONS₂(0, zeros)
ACTIVE₁(adx₁(cons₂(X, Y))) -> CONS₂(X, adx₁(Y))
ACTIVE₁(tl₁(X)) -> TL₁(active₁(X))
ACTIVE₁(incr₁(cons₂(X, Y))) -> INCR₁(Y)
ACTIVE₁(tl₁(X)) -> ACTIVE₁(X)
PROPER₁(adx₁(X)) -> PROPER₁(X)
INCR₁(mark₁(X)) -> INCR₁(X)
TL₁(ok₁(X)) -> TL₁(X)
PROPER₁(adx₁(X)) -> ADX₁(proper₁(X))
ACTIVE₁(adx₁(cons₂(X, Y))) -> ADX₁(Y)
PROPER₁(incr₁(X)) -> INCR₁(proper₁(X))
PROPER₁(incr₁(X)) -> PROPER₁(X)
ACTIVE₁(incr₁(cons₂(X, Y))) -> S₁(X)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(s₁(X)) -> S₁(proper₁(X))
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(hd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(incr₁(X)) -> ACTIVE₁(X)
ACTIVE₁(nats) -> ADX₁(zeros)
PROPER₁(hd₁(X)) -> HD₁(proper₁(X))
INCR₁(ok₁(X)) -> INCR₁(X)
PROPER₁(hd₁(X)) -> PROPER₁(X)
ADX₁(mark₁(X)) -> ADX₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(adx₁(cons₂(X, Y))) -> INCR₁(cons₂(X, adx₁(Y)))
ACTIVE₁(incr₁(X)) -> INCR₁(active₁(X))
ACTIVE₁(incr₁(cons₂(X, Y))) -> CONS₂(s₁(X), incr₁(Y))
TOP₁(ok₁(X)) -> ACTIVE₁(X)
ACTIVE₁(adx₁(X)) -> ACTIVE₁(X)
PROPER₁(tl₁(X)) -> TL₁(proper₁(X))
HD₁(mark₁(X)) -> HD₁(X)
ADX₁(ok₁(X)) -> ADX₁(X)
ACTIVE₁(hd₁(X)) -> HD₁(active₁(X))
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
PROPER₁(s₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
HD₁(ok₁(X)) -> HD₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(adx₁(X)) -> ADX₁(active₁(X))
TL₁(mark₁(X)) -> TL₁(X)
PROPER₁(tl₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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:

TOP₁(mark₁(X)) -> PROPER₁(X)
ACTIVE₁(zeros) -> CONS₂(0, zeros)
ACTIVE₁(adx₁(cons₂(X, Y))) -> CONS₂(X, adx₁(Y))
ACTIVE₁(tl₁(X)) -> TL₁(active₁(X))
ACTIVE₁(incr₁(cons₂(X, Y))) -> INCR₁(Y)
ACTIVE₁(tl₁(X)) -> ACTIVE₁(X)
PROPER₁(adx₁(X)) -> PROPER₁(X)
INCR₁(mark₁(X)) -> INCR₁(X)
TL₁(ok₁(X)) -> TL₁(X)
PROPER₁(adx₁(X)) -> ADX₁(proper₁(X))
ACTIVE₁(adx₁(cons₂(X, Y))) -> ADX₁(Y)
PROPER₁(incr₁(X)) -> INCR₁(proper₁(X))
PROPER₁(incr₁(X)) -> PROPER₁(X)
ACTIVE₁(incr₁(cons₂(X, Y))) -> S₁(X)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(s₁(X)) -> S₁(proper₁(X))
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(hd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(incr₁(X)) -> ACTIVE₁(X)
ACTIVE₁(nats) -> ADX₁(zeros)
PROPER₁(hd₁(X)) -> HD₁(proper₁(X))
INCR₁(ok₁(X)) -> INCR₁(X)
PROPER₁(hd₁(X)) -> PROPER₁(X)
ADX₁(mark₁(X)) -> ADX₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(adx₁(cons₂(X, Y))) -> INCR₁(cons₂(X, adx₁(Y)))
ACTIVE₁(incr₁(X)) -> INCR₁(active₁(X))
ACTIVE₁(incr₁(cons₂(X, Y))) -> CONS₂(s₁(X), incr₁(Y))
TOP₁(ok₁(X)) -> ACTIVE₁(X)
ACTIVE₁(adx₁(X)) -> ACTIVE₁(X)
PROPER₁(tl₁(X)) -> TL₁(proper₁(X))
HD₁(mark₁(X)) -> HD₁(X)
ADX₁(ok₁(X)) -> ADX₁(X)
ACTIVE₁(hd₁(X)) -> HD₁(active₁(X))
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
PROPER₁(s₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
HD₁(ok₁(X)) -> HD₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(adx₁(X)) -> ADX₁(active₁(X))
TL₁(mark₁(X)) -> TL₁(X)
PROPER₁(tl₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S₁(ok₁(X)) -> S₁(X)

The TRS R consists of the following rules:

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

S₁(ok₁(X)) -> S₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(S₁(x₁)) = 3·x₁
POL(ok₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)

The TRS R consists of the following rules:

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(CONS₂(x₁, x₂)) = 3·x₁ + 3·x₂
POL(ok₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

TL₁(ok₁(X)) -> TL₁(X)
TL₁(mark₁(X)) -> TL₁(X)

The TRS R consists of the following rules:

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

TL₁(ok₁(X)) -> TL₁(X)
TL₁(mark₁(X)) -> TL₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(TL₁(x₁)) = 3·x₁
POL(mark₁(x₁)) = 3 + 2·x₁
POL(ok₁(x₁)) = 3 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

HD₁(mark₁(X)) -> HD₁(X)
HD₁(ok₁(X)) -> HD₁(X)

The TRS R consists of the following rules:

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

HD₁(mark₁(X)) -> HD₁(X)
HD₁(ok₁(X)) -> HD₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(HD₁(x₁)) = 3·x₁
POL(mark₁(x₁)) = 3 + x₁
POL(ok₁(x₁)) = 3 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

INCR₁(ok₁(X)) -> INCR₁(X)
INCR₁(mark₁(X)) -> INCR₁(X)

The TRS R consists of the following rules:

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

INCR₁(ok₁(X)) -> INCR₁(X)
INCR₁(mark₁(X)) -> INCR₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(INCR₁(x₁)) = 3·x₁
POL(mark₁(x₁)) = 3 + 2·x₁
POL(ok₁(x₁)) = 3 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ADX₁(ok₁(X)) -> ADX₁(X)
ADX₁(mark₁(X)) -> ADX₁(X)

The TRS R consists of the following rules:

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

ADX₁(ok₁(X)) -> ADX₁(X)
ADX₁(mark₁(X)) -> ADX₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ADX₁(x₁)) = 3·x₁
POL(mark₁(x₁)) = 3 + 2·x₁
POL(ok₁(x₁)) = 3 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(hd₁(X)) -> PROPER₁(X)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(incr₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(adx₁(X)) -> PROPER₁(X)
PROPER₁(tl₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

PROPER₁(hd₁(X)) -> PROPER₁(X)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(incr₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(adx₁(X)) -> PROPER₁(X)
PROPER₁(tl₁(X)) -> PROPER₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROPER₁(x₁)) = 3·x₁
POL(adx₁(x₁)) = 3 + 2·x₁
POL(cons₂(x₁, x₂)) = 3 + 2·x₁ + 2·x₂
POL(hd₁(x₁)) = 3 + x₁
POL(incr₁(x₁)) = 3 + 2·x₁
POL(s₁(x₁)) = 3 + 2·x₁
POL(tl₁(x₁)) = 3 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

ACTIVE₁(hd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(tl₁(X)) -> ACTIVE₁(X)
ACTIVE₁(adx₁(X)) -> ACTIVE₁(X)
ACTIVE₁(incr₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

ACTIVE₁(hd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(tl₁(X)) -> ACTIVE₁(X)
ACTIVE₁(adx₁(X)) -> ACTIVE₁(X)
ACTIVE₁(incr₁(X)) -> ACTIVE₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE₁(x₁)) = 3·x₁
POL(adx₁(x₁)) = 3 + 2·x₁
POL(hd₁(x₁)) = 3 + x₁
POL(incr₁(x₁)) = 3 + 2·x₁
POL(tl₁(x₁)) = 3 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

TOP₁(ok₁(X)) -> TOP₁(active₁(X))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

The TRS R consists of the following rules:

active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(incr₁(cons₂(X, Y))) -> mark₁(cons₂(s₁(X), incr₁(Y)))
active₁(adx₁(cons₂(X, Y))) -> mark₁(incr₁(cons₂(X, adx₁(Y))))
active₁(hd₁(cons₂(X, Y))) -> mark₁(X)
active₁(tl₁(cons₂(X, Y))) -> mark₁(Y)
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(hd₁(X)) -> hd₁(active₁(X))
active₁(tl₁(X)) -> tl₁(active₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
hd₁(mark₁(X)) -> mark₁(hd₁(X))
tl₁(mark₁(X)) -> mark₁(tl₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(hd₁(X)) -> hd₁(proper₁(X))
proper₁(tl₁(X)) -> tl₁(proper₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
hd₁(ok₁(X)) -> ok₁(hd₁(X))
tl₁(ok₁(X)) -> ok₁(tl₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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