Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

P(a(x0), p(a(b(x1)), x2)) → P(a(b(a(x2))), p(a(a(x1)), x2))
P(a(x0), p(a(b(x1)), x2)) → P(a(a(x1)), x2)

The TRS R consists of the following rules:

p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))

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:

P(a(x0), p(a(b(x1)), x2)) → P(a(b(a(x2))), p(a(a(x1)), x2))
P(a(x0), p(a(b(x1)), x2)) → P(a(a(x1)), x2)

The TRS R consists of the following rules:

p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


P(a(x0), p(a(b(x1)), x2)) → P(a(a(x1)), x2)
The remaining pairs can at least be oriented weakly.

P(a(x0), p(a(b(x1)), x2)) → P(a(b(a(x2))), p(a(a(x1)), x2))
Used ordering: Polynomial interpretation [25]:

POL(P(x1, x2)) = x2   
POL(a(x1)) = 0   
POL(b(x1)) = 0   
POL(p(x1, x2)) = 1 + x2   

The following usable rules [17] were oriented:

p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))



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

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

P(a(x0), p(a(b(x1)), x2)) → P(a(b(a(x2))), p(a(a(x1)), x2))

The TRS R consists of the following rules:

p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


P(a(x0), p(a(b(x1)), x2)) → P(a(b(a(x2))), p(a(a(x1)), x2))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( a(x1) ) =
/0\
\0/
+
/00\
\10/
·x1

M( p(x1, x2) ) =
/0\
\1/
+
/00\
\01/
·x1+
/00\
\01/
·x2

M( b(x1) ) =
/1\
\0/
+
/10\
\00/
·x1

Tuple symbols:
M( P(x1, x2) ) = 0+
[0,0]
·x1+
[0,1]
·x2


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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

p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))

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.