Termination w.r.t. Q proof of TRS/TRCSRExIntrod_GM01

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(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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:

PROPER₁(head₁(X)) -> PROPER₁(X)
S₁(mark₁(X)) -> S₁(X)
PROPER₁(adx₁(X)) -> PROPER₁(X)
INCR₁(mark₁(X)) -> INCR₁(X)
ACTIVE₁(adx₁(cons₂(X, L))) -> ADX₁(L)
PROPER₁(incr₁(X)) -> INCR₁(proper₁(X))
PROPER₁(incr₁(X)) -> PROPER₁(X)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
ACTIVE₁(tail₁(X)) -> TAIL₁(active₁(X))
HEAD₁(mark₁(X)) -> HEAD₁(X)
ACTIVE₁(incr₁(X)) -> ACTIVE₁(X)
INCR₁(ok₁(X)) -> INCR₁(X)
ADX₁(mark₁(X)) -> ADX₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(tail₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
ACTIVE₁(adx₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(tail₁(X)) -> TAIL₁(proper₁(X))
ACTIVE₁(adx₁(X)) -> ADX₁(active₁(X))
ACTIVE₁(tail₁(X)) -> ACTIVE₁(X)
HEAD₁(ok₁(X)) -> HEAD₁(X)
ACTIVE₁(zeros) -> CONS₂(0, zeros)
TOP₁(mark₁(X)) -> PROPER₁(X)
ACTIVE₁(adx₁(cons₂(X, L))) -> CONS₂(X, adx₁(L))
TAIL₁(ok₁(X)) -> TAIL₁(X)
ACTIVE₁(incr₁(cons₂(X, L))) -> INCR₁(L)
PROPER₁(adx₁(X)) -> ADX₁(proper₁(X))
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(incr₁(cons₂(X, L))) -> S₁(X)
PROPER₁(s₁(X)) -> S₁(proper₁(X))
ACTIVE₁(head₁(X)) -> HEAD₁(active₁(X))
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(nats) -> ADX₁(zeros)
CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
TAIL₁(mark₁(X)) -> TAIL₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(head₁(X)) -> HEAD₁(proper₁(X))
ACTIVE₁(adx₁(cons₂(X, L))) -> INCR₁(cons₂(X, adx₁(L)))
ACTIVE₁(incr₁(X)) -> INCR₁(active₁(X))
ACTIVE₁(incr₁(cons₂(X, L))) -> CONS₂(s₁(X), incr₁(L))
ADX₁(ok₁(X)) -> ADX₁(X)
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)

The TRS R consists of the following rules:

active₁(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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:

PROPER₁(head₁(X)) -> PROPER₁(X)
S₁(mark₁(X)) -> S₁(X)
PROPER₁(adx₁(X)) -> PROPER₁(X)
INCR₁(mark₁(X)) -> INCR₁(X)
ACTIVE₁(adx₁(cons₂(X, L))) -> ADX₁(L)
PROPER₁(incr₁(X)) -> INCR₁(proper₁(X))
PROPER₁(incr₁(X)) -> PROPER₁(X)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
ACTIVE₁(tail₁(X)) -> TAIL₁(active₁(X))
HEAD₁(mark₁(X)) -> HEAD₁(X)
ACTIVE₁(incr₁(X)) -> ACTIVE₁(X)
INCR₁(ok₁(X)) -> INCR₁(X)
ADX₁(mark₁(X)) -> ADX₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(tail₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
ACTIVE₁(adx₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(tail₁(X)) -> TAIL₁(proper₁(X))
ACTIVE₁(adx₁(X)) -> ADX₁(active₁(X))
ACTIVE₁(tail₁(X)) -> ACTIVE₁(X)
HEAD₁(ok₁(X)) -> HEAD₁(X)
ACTIVE₁(zeros) -> CONS₂(0, zeros)
TOP₁(mark₁(X)) -> PROPER₁(X)
ACTIVE₁(adx₁(cons₂(X, L))) -> CONS₂(X, adx₁(L))
TAIL₁(ok₁(X)) -> TAIL₁(X)
ACTIVE₁(incr₁(cons₂(X, L))) -> INCR₁(L)
PROPER₁(adx₁(X)) -> ADX₁(proper₁(X))
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(incr₁(cons₂(X, L))) -> S₁(X)
PROPER₁(s₁(X)) -> S₁(proper₁(X))
ACTIVE₁(head₁(X)) -> HEAD₁(active₁(X))
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(nats) -> ADX₁(zeros)
CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
TAIL₁(mark₁(X)) -> TAIL₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(head₁(X)) -> HEAD₁(proper₁(X))
ACTIVE₁(adx₁(cons₂(X, L))) -> INCR₁(cons₂(X, adx₁(L)))
ACTIVE₁(incr₁(X)) -> INCR₁(active₁(X))
ACTIVE₁(incr₁(cons₂(X, L))) -> CONS₂(s₁(X), incr₁(L))
ADX₁(ok₁(X)) -> ADX₁(X)
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)

The TRS R consists of the following rules:

active₁(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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 22 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:

TAIL₁(mark₁(X)) -> TAIL₁(X)
TAIL₁(ok₁(X)) -> TAIL₁(X)

The TRS R consists of the following rules:

active₁(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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.

TAIL₁(mark₁(X)) -> TAIL₁(X)
TAIL₁(ok₁(X)) -> TAIL₁(X)

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

POL(TAIL₁(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

            ↳ 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(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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:

HEAD₁(ok₁(X)) -> HEAD₁(X)
HEAD₁(mark₁(X)) -> HEAD₁(X)

The TRS R consists of the following rules:

active₁(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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.

HEAD₁(ok₁(X)) -> HEAD₁(X)
HEAD₁(mark₁(X)) -> HEAD₁(X)

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

POL(HEAD₁(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

            ↳ 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(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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:

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

The TRS R consists of the following rules:

active₁(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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:

S₁(ok₁(X)) -> S₁(X)
S₁(mark₁(X)) -> S₁(X)

The TRS R consists of the following rules:

active₁(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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)
S₁(mark₁(X)) -> S₁(X)

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

POL(S₁(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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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:

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)

The TRS R consists of the following rules:

active₁(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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(mark₁(X1), X2) -> CONS₂(X1, X2)
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(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

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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:

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

The TRS R consists of the following rules:

active₁(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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(s₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(incr₁(X)) -> PROPER₁(X)
PROPER₁(tail₁(X)) -> PROPER₁(X)
PROPER₁(head₁(X)) -> PROPER₁(X)
PROPER₁(adx₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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(s₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(incr₁(X)) -> PROPER₁(X)
PROPER₁(tail₁(X)) -> PROPER₁(X)
PROPER₁(head₁(X)) -> PROPER₁(X)
PROPER₁(adx₁(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(head₁(x₁)) = 3 + 2·x₁
POL(incr₁(x₁)) = 3 + 2·x₁
POL(s₁(x₁)) = 3 + x₁
POL(tail₁(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(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
ACTIVE₁(tail₁(X)) -> ACTIVE₁(X)
ACTIVE₁(adx₁(X)) -> ACTIVE₁(X)
ACTIVE₁(incr₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
ACTIVE₁(tail₁(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(cons₂(x₁, x₂)) = 3 + 2·x₁
POL(head₁(x₁)) = 3 + x₁
POL(incr₁(x₁)) = 3 + 2·x₁
POL(s₁(x₁)) = 3 + 2·x₁
POL(tail₁(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(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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(incr₁(nil)) -> mark₁(nil)
active₁(incr₁(cons₂(X, L))) -> mark₁(cons₂(s₁(X), incr₁(L)))
active₁(adx₁(nil)) -> mark₁(nil)
active₁(adx₁(cons₂(X, L))) -> mark₁(incr₁(cons₂(X, adx₁(L))))
active₁(nats) -> mark₁(adx₁(zeros))
active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(head₁(cons₂(X, L))) -> mark₁(X)
active₁(tail₁(cons₂(X, L))) -> mark₁(L)
active₁(incr₁(X)) -> incr₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(adx₁(X)) -> adx₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
incr₁(mark₁(X)) -> mark₁(incr₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
adx₁(mark₁(X)) -> mark₁(adx₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(incr₁(X)) -> incr₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(adx₁(X)) -> adx₁(proper₁(X))
proper₁(nats) -> ok₁(nats)
proper₁(zeros) -> ok₁(zeros)
proper₁(0) -> ok₁(0)
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
incr₁(ok₁(X)) -> ok₁(incr₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
adx₁(ok₁(X)) -> ok₁(adx₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(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.