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:

incr1(nil) -> nil
incr1(cons2(X, L)) -> cons2(s1(X), incr1(L))
adx1(nil) -> nil
adx1(cons2(X, L)) -> incr1(cons2(X, adx1(L)))
nats -> adx1(zeros)
zeros -> cons2(0, zeros)
head1(cons2(X, L)) -> X
tail1(cons2(X, L)) -> L

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

incr1(nil) -> nil
incr1(cons2(X, L)) -> cons2(s1(X), incr1(L))
adx1(nil) -> nil
adx1(cons2(X, L)) -> incr1(cons2(X, adx1(L)))
nats -> adx1(zeros)
zeros -> cons2(0, zeros)
head1(cons2(X, L)) -> X
tail1(cons2(X, L)) -> L

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
QTRS
      ↳ DependencyPairsProof

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

incr1(nil) -> nil
incr1(cons2(X, L)) -> cons2(s1(X), incr1(L))
adx1(nil) -> nil
adx1(cons2(X, L)) -> incr1(cons2(X, adx1(L)))
nats -> adx1(zeros)
zeros -> cons2(0, zeros)
head1(cons2(X, L)) -> X
tail1(cons2(X, L)) -> L

The set Q consists of the following terms:

incr1(nil)
incr1(cons2(x0, x1))
adx1(nil)
adx1(cons2(x0, x1))
nats
zeros
head1(cons2(x0, x1))
tail1(cons2(x0, x1))


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

ADX1(cons2(X, L)) -> ADX1(L)
ADX1(cons2(X, L)) -> INCR1(cons2(X, adx1(L)))
NATS -> ADX1(zeros)
NATS -> ZEROS
ZEROS -> ZEROS
INCR1(cons2(X, L)) -> INCR1(L)

The TRS R consists of the following rules:

incr1(nil) -> nil
incr1(cons2(X, L)) -> cons2(s1(X), incr1(L))
adx1(nil) -> nil
adx1(cons2(X, L)) -> incr1(cons2(X, adx1(L)))
nats -> adx1(zeros)
zeros -> cons2(0, zeros)
head1(cons2(X, L)) -> X
tail1(cons2(X, L)) -> L

The set Q consists of the following terms:

incr1(nil)
incr1(cons2(x0, x1))
adx1(nil)
adx1(cons2(x0, x1))
nats
zeros
head1(cons2(x0, x1))
tail1(cons2(x0, x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

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

ADX1(cons2(X, L)) -> ADX1(L)
ADX1(cons2(X, L)) -> INCR1(cons2(X, adx1(L)))
NATS -> ADX1(zeros)
NATS -> ZEROS
ZEROS -> ZEROS
INCR1(cons2(X, L)) -> INCR1(L)

The TRS R consists of the following rules:

incr1(nil) -> nil
incr1(cons2(X, L)) -> cons2(s1(X), incr1(L))
adx1(nil) -> nil
adx1(cons2(X, L)) -> incr1(cons2(X, adx1(L)))
nats -> adx1(zeros)
zeros -> cons2(0, zeros)
head1(cons2(X, L)) -> X
tail1(cons2(X, L)) -> L

The set Q consists of the following terms:

incr1(nil)
incr1(cons2(x0, x1))
adx1(nil)
adx1(cons2(x0, x1))
nats
zeros
head1(cons2(x0, x1))
tail1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
              ↳ QDP
              ↳ QDP

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

ZEROS -> ZEROS

The TRS R consists of the following rules:

incr1(nil) -> nil
incr1(cons2(X, L)) -> cons2(s1(X), incr1(L))
adx1(nil) -> nil
adx1(cons2(X, L)) -> incr1(cons2(X, adx1(L)))
nats -> adx1(zeros)
zeros -> cons2(0, zeros)
head1(cons2(X, L)) -> X
tail1(cons2(X, L)) -> L

The set Q consists of the following terms:

incr1(nil)
incr1(cons2(x0, x1))
adx1(nil)
adx1(cons2(x0, x1))
nats
zeros
head1(cons2(x0, x1))
tail1(cons2(x0, x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP

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

INCR1(cons2(X, L)) -> INCR1(L)

The TRS R consists of the following rules:

incr1(nil) -> nil
incr1(cons2(X, L)) -> cons2(s1(X), incr1(L))
adx1(nil) -> nil
adx1(cons2(X, L)) -> incr1(cons2(X, adx1(L)))
nats -> adx1(zeros)
zeros -> cons2(0, zeros)
head1(cons2(X, L)) -> X
tail1(cons2(X, L)) -> L

The set Q consists of the following terms:

incr1(nil)
incr1(cons2(x0, x1))
adx1(nil)
adx1(cons2(x0, x1))
nats
zeros
head1(cons2(x0, x1))
tail1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


INCR1(cons2(X, L)) -> INCR1(L)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
INCR1(x1)  =  INCR1(x1)
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
cons2 > INCR1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

incr1(nil) -> nil
incr1(cons2(X, L)) -> cons2(s1(X), incr1(L))
adx1(nil) -> nil
adx1(cons2(X, L)) -> incr1(cons2(X, adx1(L)))
nats -> adx1(zeros)
zeros -> cons2(0, zeros)
head1(cons2(X, L)) -> X
tail1(cons2(X, L)) -> L

The set Q consists of the following terms:

incr1(nil)
incr1(cons2(x0, x1))
adx1(nil)
adx1(cons2(x0, x1))
nats
zeros
head1(cons2(x0, x1))
tail1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof

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

ADX1(cons2(X, L)) -> ADX1(L)

The TRS R consists of the following rules:

incr1(nil) -> nil
incr1(cons2(X, L)) -> cons2(s1(X), incr1(L))
adx1(nil) -> nil
adx1(cons2(X, L)) -> incr1(cons2(X, adx1(L)))
nats -> adx1(zeros)
zeros -> cons2(0, zeros)
head1(cons2(X, L)) -> X
tail1(cons2(X, L)) -> L

The set Q consists of the following terms:

incr1(nil)
incr1(cons2(x0, x1))
adx1(nil)
adx1(cons2(x0, x1))
nats
zeros
head1(cons2(x0, x1))
tail1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ADX1(cons2(X, L)) -> ADX1(L)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ADX1(x1)  =  ADX1(x1)
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
cons2 > ADX1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof

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

incr1(nil) -> nil
incr1(cons2(X, L)) -> cons2(s1(X), incr1(L))
adx1(nil) -> nil
adx1(cons2(X, L)) -> incr1(cons2(X, adx1(L)))
nats -> adx1(zeros)
zeros -> cons2(0, zeros)
head1(cons2(X, L)) -> X
tail1(cons2(X, L)) -> L

The set Q consists of the following terms:

incr1(nil)
incr1(cons2(x0, x1))
adx1(nil)
adx1(cons2(x0, x1))
nats
zeros
head1(cons2(x0, x1))
tail1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.