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:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(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:

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

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(x0, x1))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(x0, x1))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

INCR(cons(X, L)) → INCR(L)

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(x0, x1))

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.


INCR(cons(X, L)) → INCR(L)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
INCR(x1)  =  x1
cons(x1, x2)  =  cons(x2)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof

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

ADX(cons(X, L)) → ADX(L)

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(x0, x1))

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.


ADX(cons(X, L)) → ADX(L)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
ADX(x1)  =  x1
cons(x1, x2)  =  cons(x2)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(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.