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:

p2(a1(x0), p2(b1(a1(x1)), x2)) -> p2(x1, p2(a1(b1(a1(x1))), x2))
a1(b1(a1(x0))) -> b1(a1(b1(x0)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

p2(a1(x0), p2(b1(a1(x1)), x2)) -> p2(x1, p2(a1(b1(a1(x1))), x2))
a1(b1(a1(x0))) -> b1(a1(b1(x0)))

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:

P2(a1(x0), p2(b1(a1(x1)), x2)) -> P2(x1, p2(a1(b1(a1(x1))), x2))
P2(a1(x0), p2(b1(a1(x1)), x2)) -> P2(a1(b1(a1(x1))), x2)
A1(b1(a1(x0))) -> A1(b1(x0))
P2(a1(x0), p2(b1(a1(x1)), x2)) -> A1(b1(a1(x1)))

The TRS R consists of the following rules:

p2(a1(x0), p2(b1(a1(x1)), x2)) -> p2(x1, p2(a1(b1(a1(x1))), x2))
a1(b1(a1(x0))) -> b1(a1(b1(x0)))

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:

P2(a1(x0), p2(b1(a1(x1)), x2)) -> P2(x1, p2(a1(b1(a1(x1))), x2))
P2(a1(x0), p2(b1(a1(x1)), x2)) -> P2(a1(b1(a1(x1))), x2)
A1(b1(a1(x0))) -> A1(b1(x0))
P2(a1(x0), p2(b1(a1(x1)), x2)) -> A1(b1(a1(x1)))

The TRS R consists of the following rules:

p2(a1(x0), p2(b1(a1(x1)), x2)) -> p2(x1, p2(a1(b1(a1(x1))), x2))
a1(b1(a1(x0))) -> b1(a1(b1(x0)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

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

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

A1(b1(a1(x0))) -> A1(b1(x0))

The TRS R consists of the following rules:

p2(a1(x0), p2(b1(a1(x1)), x2)) -> p2(x1, p2(a1(b1(a1(x1))), x2))
a1(b1(a1(x0))) -> b1(a1(b1(x0)))

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.


A1(b1(a1(x0))) -> A1(b1(x0))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( A1(x1) ) = max{0, x1 - 3}


POL( b1(x1) ) = x1 + 1


POL( a1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



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

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

p2(a1(x0), p2(b1(a1(x1)), x2)) -> p2(x1, p2(a1(b1(a1(x1))), x2))
a1(b1(a1(x0))) -> b1(a1(b1(x0)))

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

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

P2(a1(x0), p2(b1(a1(x1)), x2)) -> P2(x1, p2(a1(b1(a1(x1))), x2))
P2(a1(x0), p2(b1(a1(x1)), x2)) -> P2(a1(b1(a1(x1))), x2)

The TRS R consists of the following rules:

p2(a1(x0), p2(b1(a1(x1)), x2)) -> p2(x1, p2(a1(b1(a1(x1))), x2))
a1(b1(a1(x0))) -> b1(a1(b1(x0)))

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