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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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

POL(P2(x1, x2)) = 2·x2   
POL(a1(x1)) = 0   
POL(b1(x1)) = 0   
POL(p2(x1, x2)) = 1 + x2   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP

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

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

The TRS R consists of the following rules:

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

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