Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPOrderProof (⇔)
↳4 QDP
↳5 ForwardInstantiation (⇔)
↳6 QDP
↳7 SemLabProof (⇐)
↳8 QDP
↳9 DependencyGraphProof (⇔)
↳10 AND
  ↳11 QDP
    ↳12 QDPOrderProof (⇔)
    ↳13 QDP
    ↳14 PisEmptyProof (⇔)
    ↳15 TRUE
  ↳16 QDP
    ↳17 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 PisEmptyProof (⇔)
    ↳20 TRUE

(0) Obligation:

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

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

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

The TRS R consists of the following rules:

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

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


P(a(x0), p(x1, p(x2, x3))) → P(x0, p(a(x3), x3))
P(a(x0), p(x1, p(x2, x3))) → P(a(x3), x3)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented:

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

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(5) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule P(a(x0), p(x1, p(x2, x3))) → P(x1, p(x0, p(a(x3), x3))) we obtained the following new rules [LPAR04]:

P(a(x0), p(a(y_0), p(x2, x3))) → P(a(y_0), p(x0, p(a(x3), x3)))

(6) Obligation:

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

P(a(x0), p(a(y_0), p(x2, x3))) → P(a(y_0), p(x0, p(a(x3), x3)))

The TRS R consists of the following rules:

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

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

(7) SemLabProof (SOUND transformation)

We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.P: 0
a: 1
p: 0
By semantic labelling [SEMLAB] we obtain the following labelled TRS.

(8) Obligation:

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

P.1-0(a.0(x0), p.1-0(a.0(y_0), p.0-0(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.0(y_0), p.0-1(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.0(y_0), p.1-0(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.0(y_0), p.1-1(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.0-0(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.0-1(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.1-0(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.1-1(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.0-0(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.0-1(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.1-0(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.1-1(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.0-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.0-1(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-1(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))

The TRS R consists of the following rules:

p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 13 less nodes.

(10) Complex Obligation (AND)

(11) Obligation:

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

P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-1(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))

The TRS R consists of the following rules:

p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-1(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented: none

(13) Obligation:

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

p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))

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

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) TRUE

(16) Obligation:

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

P.1-0(a.1(x0), p.1-0(a.1(y_0), p.0-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))

The TRS R consists of the following rules:

p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


P.1-0(a.1(x0), p.1-0(a.1(y_0), p.0-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented:

p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))

(18) Obligation:

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

p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE