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(p(b(a(x0)), x1), p(x2, x3)) → p(p(x3, a(x2)), p(b(a(x1)), b(x0)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

p(p(b(a(x0)), x1), p(x2, x3)) → p(p(x3, a(x2)), p(b(a(x1)), b(x0)))

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(p(b(a(x0)), x1), p(x2, x3)) → P(x3, a(x2))
P(p(b(a(x0)), x1), p(x2, x3)) → P(b(a(x1)), b(x0))
P(p(b(a(x0)), x1), p(x2, x3)) → P(p(x3, a(x2)), p(b(a(x1)), b(x0)))

The TRS R consists of the following rules:

p(p(b(a(x0)), x1), p(x2, x3)) → p(p(x3, a(x2)), p(b(a(x1)), b(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:

P(p(b(a(x0)), x1), p(x2, x3)) → P(x3, a(x2))
P(p(b(a(x0)), x1), p(x2, x3)) → P(b(a(x1)), b(x0))
P(p(b(a(x0)), x1), p(x2, x3)) → P(p(x3, a(x2)), p(b(a(x1)), b(x0)))

The TRS R consists of the following rules:

p(p(b(a(x0)), x1), p(x2, x3)) → p(p(x3, a(x2)), p(b(a(x1)), b(x0)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof

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

P(p(b(a(x0)), x1), p(x2, x3)) → P(p(x3, a(x2)), p(b(a(x1)), b(x0)))

The TRS R consists of the following rules:

p(p(b(a(x0)), x1), p(x2, x3)) → p(p(x3, a(x2)), p(b(a(x1)), b(x0)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

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

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

P(p(b(a(x0)), x1), p(x2, x3)) → P(p(x3, a(x2)), p(b(a(x1)), b(x0)))

R is empty.
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(p(b(a(x0)), x1), p(x2, x3)) → P(p(x3, a(x2)), p(b(a(x1)), b(x0)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( a(x1) ) =
/1\
\1/
+
/01\
\11/
·x1

M( p(x1, x2) ) =
/0\
\0/
+
/00\
\11/
·x1+
/11\
\00/
·x2

M( b(x1) ) =
/0\
\0/
+
/11\
\11/
·x1

Tuple symbols:
M( P(x1, x2) ) = 0+
[0,1]
·x1+
[1,0]
·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: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
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.