(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
q(0(x)) → p(p(s(s(0(s(s(s(s(x)))))))))
q(s(x)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x)))))))))))))
r(0(x)) → p(s(p(s(0(p(p(p(s(s(s(x)))))))))))
r(s(x)) → p(s(p(s(s(q(p(s(p(s(x))))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
p(0(x)) → 0(s(s(s(x))))
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0(x1)) = x1
POL(p(x1)) = x1
POL(q(x1)) = 1 + x1
POL(r(x1)) = 1 + x1
POL(s(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
q(0(x)) → p(p(s(s(0(s(s(s(s(x)))))))))
r(0(x)) → p(s(p(s(0(p(p(p(s(s(s(x)))))))))))
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
q(s(x)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x)))))))))))))
r(s(x)) → p(s(p(s(s(q(p(s(p(s(x))))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
p(0(x)) → 0(s(s(s(x))))
Q is empty.
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
Q(s(x)) → P(p(s(s(s(s(s(s(r(p(p(s(s(x)))))))))))))
Q(s(x)) → P(s(s(s(s(s(s(r(p(p(s(s(x))))))))))))
Q(s(x)) → R(p(p(s(s(x)))))
Q(s(x)) → P(p(s(s(x))))
Q(s(x)) → P(s(s(x)))
R(s(x)) → P(s(p(s(s(q(p(s(p(s(x))))))))))
R(s(x)) → P(s(s(q(p(s(p(s(x))))))))
R(s(x)) → Q(p(s(p(s(x)))))
R(s(x)) → P(s(p(s(x))))
R(s(x)) → P(s(x))
P(p(s(x))) → P(x)
The TRS R consists of the following rules:
q(s(x)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x)))))))))))))
r(s(x)) → p(s(p(s(s(q(p(s(p(s(x))))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
p(0(x)) → 0(s(s(s(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 8 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P(p(s(x))) → P(x)
The TRS R consists of the following rules:
q(s(x)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x)))))))))))))
r(s(x)) → p(s(p(s(s(q(p(s(p(s(x))))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
p(0(x)) → 0(s(s(s(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P(p(s(x))) → P(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- P(p(s(x))) → P(x)
The graph contains the following edges 1 > 1
(11) YES
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
Q(s(x)) → R(p(p(s(s(x)))))
R(s(x)) → Q(p(s(p(s(x)))))
The TRS R consists of the following rules:
q(s(x)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x)))))))))))))
r(s(x)) → p(s(p(s(s(q(p(s(p(s(x))))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
p(0(x)) → 0(s(s(s(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
Q(s(x)) → R(p(p(s(s(x)))))
R(s(x)) → Q(p(s(p(s(x)))))
The TRS R consists of the following rules:
q(s(x)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x)))))))))))))
r(s(x)) → p(s(p(s(s(q(p(s(p(s(x))))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
p(0(x)) → 0(s(s(s(x))))
The set Q consists of the following terms:
q(s(x0))
r(s(x0))
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
(15) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
Q(s(x)) → R(p(p(s(s(x)))))
R(s(x)) → Q(p(s(p(s(x)))))
The TRS R consists of the following rules:
p(s(x)) → x
p(0(x)) → 0(s(s(s(x))))
The set Q consists of the following terms:
q(s(x0))
r(s(x0))
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
(17) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
q(s(x0))
r(s(x0))
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
Q(s(x)) → R(p(p(s(s(x)))))
R(s(x)) → Q(p(s(p(s(x)))))
The TRS R consists of the following rules:
p(s(x)) → x
p(0(x)) → 0(s(s(s(x))))
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
(19) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
Q(s(x)) → R(p(p(s(s(x)))))
R(s(x)) → Q(p(s(p(s(x)))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( Q(x1) ) = max{0, 2x1 - 2} |
| POL( R(x1) ) = max{0, 2x1 - 2} |
| POL( p(x1) ) = max{0, x1 - 2} |
| POL( 0(x1) ) = max{0, -2} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
p(s(x)) → x
p(0(x)) → 0(s(s(s(x))))
(20) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
p(s(x)) → x
p(0(x)) → 0(s(s(s(x))))
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
(21) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(22) YES