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(x1), p2(a1(x2), x3))) -> p2(x2, p2(a1(a1(x0)), p2(b1(x1), x3)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

p2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> p2(x2, p2(a1(a1(x0)), p2(b1(x1), x3)))

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(x1), p2(a1(x2), x3))) -> P2(x2, p2(a1(a1(x0)), p2(b1(x1), x3)))
P2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> P2(a1(a1(x0)), p2(b1(x1), x3))
P2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> P2(b1(x1), x3)

The TRS R consists of the following rules:

p2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> p2(x2, p2(a1(a1(x0)), p2(b1(x1), x3)))

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(x1), p2(a1(x2), x3))) -> P2(x2, p2(a1(a1(x0)), p2(b1(x1), x3)))
P2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> P2(a1(a1(x0)), p2(b1(x1), x3))
P2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> P2(b1(x1), x3)

The TRS R consists of the following rules:

p2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> p2(x2, p2(a1(a1(x0)), p2(b1(x1), x3)))

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

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

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

P2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> P2(x2, p2(a1(a1(x0)), p2(b1(x1), x3)))
P2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> P2(a1(a1(x0)), p2(b1(x1), x3))

The TRS R consists of the following rules:

p2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> p2(x2, p2(a1(a1(x0)), p2(b1(x1), x3)))

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.


P2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> P2(a1(a1(x0)), p2(b1(x1), x3))
The remaining pairs can at least be oriented weakly.

P2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> P2(x2, p2(a1(a1(x0)), p2(b1(x1), x3)))
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( P2(x1, x2) ) = max{0, x1 + x2 - 1}


POL( a1(x1) ) = x1


POL( b1(x1) ) = 1


POL( p2(x1, x2) ) = x1 + x2 + 1



The following usable rules [14] were oriented:

p2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> p2(x2, p2(a1(a1(x0)), p2(b1(x1), x3)))



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

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

P2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> P2(x2, p2(a1(a1(x0)), p2(b1(x1), x3)))

The TRS R consists of the following rules:

p2(a1(x0), p2(b1(x1), p2(a1(x2), x3))) -> p2(x2, p2(a1(a1(x0)), p2(b1(x1), x3)))

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