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.

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 contains 9 SCCs with 22 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TAIL1(ok1(X)) -> TAIL1(X)
Used argument filtering: TAIL1(x1)  =  x1
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

TAIL1(mark1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TAIL1(mark1(X)) -> TAIL1(X)
Used argument filtering: TAIL1(x1)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

HEAD1(mark1(X)) -> HEAD1(X)
Used argument filtering: HEAD1(x1)  =  x1
ok1(x1)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

HEAD1(ok1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

HEAD1(ok1(X)) -> HEAD1(X)
Used argument filtering: HEAD1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ADX1(mark1(X)) -> ADX1(X)
Used argument filtering: ADX1(x1)  =  x1
ok1(x1)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

ADX1(ok1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ADX1(ok1(X)) -> ADX1(X)
Used argument filtering: ADX1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S1(mark1(X)) -> S1(X)
Used argument filtering: S1(x1)  =  x1
ok1(x1)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ 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(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S1(ok1(X)) -> S1(X)
Used argument filtering: S1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
Used argument filtering: CONS2(x1, x2)  =  x2
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

CONS2(mark1(X1), 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
Used argument filtering: CONS2(x1, x2)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

INCR1(mark1(X)) -> INCR1(X)
Used argument filtering: INCR1(x1)  =  x1
ok1(x1)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

INCR1(ok1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

INCR1(ok1(X)) -> INCR1(X)
Used argument filtering: INCR1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
Used argument filtering: PROPER1(x1)  =  x1
s1(x1)  =  x1
cons2(x1, x2)  =  cons2(x1, x2)
incr1(x1)  =  x1
tail1(x1)  =  x1
head1(x1)  =  x1
adx1(x1)  =  x1
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

PROPER1(s1(X)) -> PROPER1(X)
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(adx1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1)  =  x1
s1(x1)  =  x1
incr1(x1)  =  x1
tail1(x1)  =  x1
head1(x1)  =  x1
adx1(x1)  =  adx1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(tail1(X)) -> PROPER1(X)
PROPER1(head1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(head1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1)  =  x1
s1(x1)  =  x1
incr1(x1)  =  x1
tail1(x1)  =  x1
head1(x1)  =  head1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
QDP
                        ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(tail1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(tail1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1)  =  x1
s1(x1)  =  x1
incr1(x1)  =  x1
tail1(x1)  =  tail1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
                      ↳ QDP
                        ↳ QDPAfsSolverProof
QDP
                            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(incr1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(incr1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1)  =  x1
s1(x1)  =  x1
incr1(x1)  =  incr1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
                      ↳ QDP
                        ↳ QDPAfsSolverProof
                          ↳ QDP
                            ↳ QDPAfsSolverProof
QDP
                                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

PROPER1(s1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(s1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
                      ↳ QDP
                        ↳ QDPAfsSolverProof
                          ↳ QDP
                            ↳ QDPAfsSolverProof
                              ↳ QDP
                                ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(incr1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1)  =  x1
head1(x1)  =  x1
cons2(x1, x2)  =  x1
s1(x1)  =  x1
tail1(x1)  =  x1
adx1(x1)  =  x1
incr1(x1)  =  incr1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ 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)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(adx1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1)  =  x1
head1(x1)  =  x1
cons2(x1, x2)  =  x1
s1(x1)  =  x1
tail1(x1)  =  x1
adx1(x1)  =  adx1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof
          ↳ 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)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(tail1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1)  =  x1
head1(x1)  =  x1
cons2(x1, x2)  =  x1
s1(x1)  =  x1
tail1(x1)  =  tail1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
QDP
                        ↳ QDPAfsSolverProof
          ↳ 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)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(s1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1)  =  x1
head1(x1)  =  x1
cons2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
                      ↳ QDP
                        ↳ QDPAfsSolverProof
QDP
                            ↳ QDPAfsSolverProof
          ↳ QDP

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

ACTIVE1(head1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
Used argument filtering: ACTIVE1(x1)  =  x1
head1(x1)  =  x1
cons2(x1, x2)  =  cons1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
                      ↳ QDP
                        ↳ QDPAfsSolverProof
                          ↳ QDP
                            ↳ QDPAfsSolverProof
QDP
                                ↳ QDPAfsSolverProof
          ↳ QDP

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

ACTIVE1(head1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(head1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1)  =  x1
head1(x1)  =  head1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
                      ↳ QDP
                        ↳ QDPAfsSolverProof
                          ↳ QDP
                            ↳ QDPAfsSolverProof
                              ↳ QDP
                                ↳ QDPAfsSolverProof
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.