(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(s(x)) → s(s(s(p(s(b(p(p(s(s(x))))))))))
b(s(x)) → s(s(s(p(p(s(s(c(p(s(p(s(x))))))))))))
c(s(x)) → p(s(p(s(a(p(s(p(s(x)))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
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:
A(s(x)) → P(s(b(p(p(s(s(x)))))))
A(s(x)) → B(p(p(s(s(x)))))
A(s(x)) → P(p(s(s(x))))
A(s(x)) → P(s(s(x)))
B(s(x)) → P(p(s(s(c(p(s(p(s(x)))))))))
B(s(x)) → P(s(s(c(p(s(p(s(x))))))))
B(s(x)) → C(p(s(p(s(x)))))
B(s(x)) → P(s(p(s(x))))
B(s(x)) → P(s(x))
C(s(x)) → P(s(p(s(a(p(s(p(s(x)))))))))
C(s(x)) → P(s(a(p(s(p(s(x)))))))
C(s(x)) → A(p(s(p(s(x)))))
C(s(x)) → P(s(p(s(x))))
C(s(x)) → P(s(x))
P(p(s(x))) → P(x)
The TRS R consists of the following rules:
a(s(x)) → s(s(s(p(s(b(p(p(s(s(x))))))))))
b(s(x)) → s(s(s(p(p(s(s(c(p(s(p(s(x))))))))))))
c(s(x)) → p(s(p(s(a(p(s(p(s(x)))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
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 11 less nodes.
(4) Complex Obligation (AND)
(5) 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:
a(s(x)) → s(s(s(p(s(b(p(p(s(s(x))))))))))
b(s(x)) → s(s(s(p(p(s(s(c(p(s(p(s(x))))))))))))
c(s(x)) → p(s(p(s(a(p(s(p(s(x)))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) 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.
(7) 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.
(8) 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
(9) YES
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(s(x)) → B(p(p(s(s(x)))))
B(s(x)) → C(p(s(p(s(x)))))
C(s(x)) → A(p(s(p(s(x)))))
The TRS R consists of the following rules:
a(s(x)) → s(s(s(p(s(b(p(p(s(s(x))))))))))
b(s(x)) → s(s(s(p(p(s(s(c(p(s(p(s(x))))))))))))
c(s(x)) → p(s(p(s(a(p(s(p(s(x)))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(s(x)) → B(p(p(s(s(x)))))
B(s(x)) → C(p(s(p(s(x)))))
C(s(x)) → A(p(s(p(s(x)))))
The TRS R consists of the following rules:
a(s(x)) → s(s(s(p(s(b(p(p(s(s(x))))))))))
b(s(x)) → s(s(s(p(p(s(s(c(p(s(p(s(x))))))))))))
c(s(x)) → p(s(p(s(a(p(s(p(s(x)))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
The set Q consists of the following terms:
a(s(x0))
b(s(x0))
c(s(x0))
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(13) 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.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(s(x)) → B(p(p(s(s(x)))))
B(s(x)) → C(p(s(p(s(x)))))
C(s(x)) → A(p(s(p(s(x)))))
The TRS R consists of the following rules:
p(s(x)) → x
The set Q consists of the following terms:
a(s(x0))
b(s(x0))
c(s(x0))
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(15) 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].
a(s(x0))
b(s(x0))
c(s(x0))
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(s(x)) → B(p(p(s(s(x)))))
B(s(x)) → C(p(s(p(s(x)))))
C(s(x)) → A(p(s(p(s(x)))))
The TRS R consists of the following rules:
p(s(x)) → x
The set Q consists of the following terms:
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(17) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
C(s(x)) → A(p(s(p(s(x)))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( A(x1) ) = max{0, 2x1 - 2} |
| POL( B(x1) ) = max{0, 2x1 - 2} |
| POL( C(x1) ) = max{0, 2x1 - 1} |
| POL( p(x1) ) = max{0, x1 - 1} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
p(s(x)) → x
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(s(x)) → B(p(p(s(s(x)))))
B(s(x)) → C(p(s(p(s(x)))))
The TRS R consists of the following rules:
p(s(x)) → x
The set Q consists of the following terms:
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(19) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
(20) TRUE