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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: POLO with Polynomial interpretation [25]:

POL(P(x1, x2)) = 2·x1 + x2   
POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(p(x1, x2)) = 1 + 2·x1 + x2   



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

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

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

The TRS R consists of the following rules:

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

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(b(x1), p(a(x2), x3))) → P(x2, p(a(a(x0)), p(b(x1), x3)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( a(x1) ) =
/0\
\0/
+
/11\
\00/
·x1

M( p(x1, x2) ) =
/0\
\0/
+
/10\
\11/
·x1+
/10\
\10/
·x2

M( b(x1) ) =
/0\
\1/
+
/10\
\10/
·x1

Tuple symbols:
M( P(x1, x2) ) = 0+
[1,0]
·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(b(x1), p(a(x2), x3))) → p(x2, p(a(a(x0)), p(b(x1), x3)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

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

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

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.