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 -> adx1(zeros)
zeros -> cons2(0, zeros)
incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
adx1(cons2(X, Y)) -> incr1(cons2(X, adx1(Y)))
hd1(cons2(X, Y)) -> X
tl1(cons2(X, Y)) -> Y

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

nats -> adx1(zeros)
zeros -> cons2(0, zeros)
incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
adx1(cons2(X, Y)) -> incr1(cons2(X, adx1(Y)))
hd1(cons2(X, Y)) -> X
tl1(cons2(X, Y)) -> Y

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:

nats -> adx1(zeros)
zeros -> cons2(0, zeros)
incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
adx1(cons2(X, Y)) -> incr1(cons2(X, adx1(Y)))
hd1(cons2(X, Y)) -> X
tl1(cons2(X, Y)) -> Y

The set Q consists of the following terms:

nats
zeros
incr1(cons2(x0, x1))
adx1(cons2(x0, x1))
hd1(cons2(x0, x1))
tl1(cons2(x0, x1))


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

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

The TRS R consists of the following rules:

nats -> adx1(zeros)
zeros -> cons2(0, zeros)
incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
adx1(cons2(X, Y)) -> incr1(cons2(X, adx1(Y)))
hd1(cons2(X, Y)) -> X
tl1(cons2(X, Y)) -> Y

The set Q consists of the following terms:

nats
zeros
incr1(cons2(x0, x1))
adx1(cons2(x0, x1))
hd1(cons2(x0, x1))
tl1(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, Y)) -> ADX1(Y)
ADX1(cons2(X, Y)) -> INCR1(cons2(X, adx1(Y)))
NATS -> ADX1(zeros)
NATS -> ZEROS
ZEROS -> ZEROS
INCR1(cons2(X, Y)) -> INCR1(Y)

The TRS R consists of the following rules:

nats -> adx1(zeros)
zeros -> cons2(0, zeros)
incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
adx1(cons2(X, Y)) -> incr1(cons2(X, adx1(Y)))
hd1(cons2(X, Y)) -> X
tl1(cons2(X, Y)) -> Y

The set Q consists of the following terms:

nats
zeros
incr1(cons2(x0, x1))
adx1(cons2(x0, x1))
hd1(cons2(x0, x1))
tl1(cons2(x0, x1))

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

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

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

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

The TRS R consists of the following rules:

nats -> adx1(zeros)
zeros -> cons2(0, zeros)
incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
adx1(cons2(X, Y)) -> incr1(cons2(X, adx1(Y)))
hd1(cons2(X, Y)) -> X
tl1(cons2(X, Y)) -> Y

The set Q consists of the following terms:

nats
zeros
incr1(cons2(x0, x1))
adx1(cons2(x0, x1))
hd1(cons2(x0, x1))
tl1(cons2(x0, x1))

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(cons2(X, Y)) -> INCR1(Y)
Used argument filtering: INCR1(x1)  =  x1
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial


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

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

nats -> adx1(zeros)
zeros -> cons2(0, zeros)
incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
adx1(cons2(X, Y)) -> incr1(cons2(X, adx1(Y)))
hd1(cons2(X, Y)) -> X
tl1(cons2(X, Y)) -> Y

The set Q consists of the following terms:

nats
zeros
incr1(cons2(x0, x1))
adx1(cons2(x0, x1))
hd1(cons2(x0, x1))
tl1(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
                ↳ QDPAfsSolverProof
              ↳ QDP

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

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

The TRS R consists of the following rules:

nats -> adx1(zeros)
zeros -> cons2(0, zeros)
incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
adx1(cons2(X, Y)) -> incr1(cons2(X, adx1(Y)))
hd1(cons2(X, Y)) -> X
tl1(cons2(X, Y)) -> Y

The set Q consists of the following terms:

nats
zeros
incr1(cons2(x0, x1))
adx1(cons2(x0, x1))
hd1(cons2(x0, x1))
tl1(cons2(x0, x1))

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(cons2(X, Y)) -> ADX1(Y)
Used argument filtering: ADX1(x1)  =  x1
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial


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

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

nats -> adx1(zeros)
zeros -> cons2(0, zeros)
incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
adx1(cons2(X, Y)) -> incr1(cons2(X, adx1(Y)))
hd1(cons2(X, Y)) -> X
tl1(cons2(X, Y)) -> Y

The set Q consists of the following terms:

nats
zeros
incr1(cons2(x0, x1))
adx1(cons2(x0, x1))
hd1(cons2(x0, x1))
tl1(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

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

ZEROS -> ZEROS

The TRS R consists of the following rules:

nats -> adx1(zeros)
zeros -> cons2(0, zeros)
incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
adx1(cons2(X, Y)) -> incr1(cons2(X, adx1(Y)))
hd1(cons2(X, Y)) -> X
tl1(cons2(X, Y)) -> Y

The set Q consists of the following terms:

nats
zeros
incr1(cons2(x0, x1))
adx1(cons2(x0, x1))
hd1(cons2(x0, x1))
tl1(cons2(x0, x1))

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