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
  ↳ 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

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:

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

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:

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

Q is empty.
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
  ↳ 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

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

↳ 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

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(cons2(X, L)) -> INCR1(L)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( cons2(x1, x2) ) = 3x1 + x2 + 1


POL( INCR1(x1) ) = 2x1 + 2



The following usable rules [14] were oriented: none



↳ 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

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

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

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(cons2(X, L)) -> ADX1(L)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( cons2(x1, x2) ) = 3x1 + x2 + 1


POL( ADX1(x1) ) = 2x1 + 2



The following usable rules [14] were oriented: none



↳ 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

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.