(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
R(x) → r(x)
r(p(x)) → p(p(r(P(x))))
r(r(x)) → x
r(P(P(x))) → P(P(r(x)))
p(P(x)) → x
P(p(x)) → x
r(R(x)) → x
R(r(x)) → x
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(P(x1)) = x1
POL(R(x1)) = 1 + x1
POL(p(x1)) = x1
POL(r(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
R(x) → r(x)
r(R(x)) → x
R(r(x)) → x
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
r(p(x)) → p(p(r(P(x))))
r(r(x)) → x
r(P(P(x))) → P(P(r(x)))
p(P(x)) → x
P(p(x)) → x
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(P(x1)) = x1
POL(p(x1)) = x1
POL(r(x1)) = 1 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
r(r(x)) → x
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
r(p(x)) → p(p(r(P(x))))
r(P(P(x))) → P(P(r(x)))
p(P(x)) → x
P(p(x)) → x
Q is empty.
(5) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
R(p(x)) → P1(p(r(P(x))))
R(p(x)) → P1(r(P(x)))
R(p(x)) → R(P(x))
R(p(x)) → P2(x)
R(P(P(x))) → P2(P(r(x)))
R(P(P(x))) → P2(r(x))
R(P(P(x))) → R(x)
The TRS R consists of the following rules:
r(p(x)) → p(p(r(P(x))))
r(P(P(x))) → P(P(r(x)))
p(P(x)) → x
P(p(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
R(P(P(x))) → R(x)
R(p(x)) → R(P(x))
The TRS R consists of the following rules:
r(p(x)) → p(p(r(P(x))))
r(P(P(x))) → P(P(r(x)))
p(P(x)) → x
P(p(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
R(P(P(x))) → R(x)
R(p(x)) → R(P(x))
The TRS R consists of the following rules:
P(p(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
R(P(P(x))) → R(x)
R(p(x)) → R(P(x))
Strictly oriented rules of the TRS R:
P(p(x)) → x
Used ordering: Polynomial interpretation [POLO]:
POL(P(x1)) = 1 + 3·x1
POL(R(x1)) = 2·x1
POL(p(x1)) = 2 + 3·x1
(12) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(14) YES