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:

nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ADX(cons(X, Y)) → ADX(Y)
ZEROS → ZEROS
NATS → ZEROS
ADX(cons(X, Y)) → INCR(cons(X, adx(Y)))
NATS → ADX(zeros)
INCR(cons(X, Y)) → INCR(Y)

The TRS R consists of the following rules:

nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

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:

ADX(cons(X, Y)) → ADX(Y)
ZEROS → ZEROS
NATS → ZEROS
ADX(cons(X, Y)) → INCR(cons(X, adx(Y)))
NATS → ADX(zeros)
INCR(cons(X, Y)) → INCR(Y)

The TRS R consists of the following rules:

nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

INCR(cons(X, Y)) → INCR(Y)

The TRS R consists of the following rules:

nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


INCR(cons(X, Y)) → INCR(Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x1, x2)) = 1/4 + (7/2)x_2   
POL(INCR(x1)) = (2)x_1   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

ADX(cons(X, Y)) → ADX(Y)

The TRS R consists of the following rules:

nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


ADX(cons(X, Y)) → ADX(Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x1, x2)) = 1/4 + (7/2)x_2   
POL(ADX(x1)) = (2)x_1   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] 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:

nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

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

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

ZEROS → ZEROS

The TRS R consists of the following rules:

nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

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