Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

A(b(a(x0))) → A(b(x0))

The TRS R consists of the following rules:

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

A(b(a(x0))) → A(b(x0))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A(x1)  =  A(x1)
b(x1)  =  b(x1)
a(x1)  =  a(x1)
p(x1, x2)  =  p(x2)

Lexicographic path order with status [LPO].
Precedence:

A₁ > b₁ > p₁
a₁ > b₁ > p₁

Status:

a₁: [1]
A₁: [1]
p₁: [1]
b₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

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

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

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
P(x1, x2)  =  P(x2)
a(x1)  =  x1
p(x1, x2)  =  p(x2)
b(x1)  =  b

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

P₁: [1]
p₁: [1]
b: []

The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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

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

The TRS R consists of the following rules:

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

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