(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(f, 0) → a(s, 0)
a(d, 0) → 0
a(d, a(s, x)) → a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) → a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) → x
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 2
POL(a(x1, x2)) = 2·x1 + x2
POL(d) = 0
POL(f) = 1
POL(p) = 0
POL(s) = 0
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
a(f, 0) → a(s, 0)
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(d, 0) → 0
a(d, a(s, x)) → a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) → a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) → x
Q is empty.
(3) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(d, 0) → 0
a(d, a(s, x)) → a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) → a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) → x
The set Q consists of the following terms:
a(d, 0)
a(d, a(s, x0))
a(f, a(s, x0))
a(p, a(s, x0))
(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:
A(d, a(s, x)) → A(s, a(s, a(d, a(p, a(s, x)))))
A(d, a(s, x)) → A(s, a(d, a(p, a(s, x))))
A(d, a(s, x)) → A(d, a(p, a(s, x)))
A(d, a(s, x)) → A(p, a(s, x))
A(f, a(s, x)) → A(d, a(f, a(p, a(s, x))))
A(f, a(s, x)) → A(f, a(p, a(s, x)))
A(f, a(s, x)) → A(p, a(s, x))
The TRS R consists of the following rules:
a(d, 0) → 0
a(d, a(s, x)) → a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) → a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) → x
The set Q consists of the following terms:
a(d, 0)
a(d, a(s, x0))
a(f, a(s, x0))
a(p, a(s, x0))
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 5 less nodes.
(8) Complex Obligation (AND)
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(d, a(s, x)) → A(d, a(p, a(s, x)))
The TRS R consists of the following rules:
a(d, 0) → 0
a(d, a(s, x)) → a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) → a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) → x
The set Q consists of the following terms:
a(d, 0)
a(d, a(s, x0))
a(f, a(s, x0))
a(p, a(s, x0))
We have to consider all minimal (P,Q,R)-chains.
(10) 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.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(d, a(s, x)) → A(d, a(p, a(s, x)))
The TRS R consists of the following rules:
a(p, a(s, x)) → x
The set Q consists of the following terms:
a(d, 0)
a(d, a(s, x0))
a(f, a(s, x0))
a(p, a(s, x0))
We have to consider all minimal (P,Q,R)-chains.
(12) ATransformationProof (EQUIVALENT transformation)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
d1(s(x)) → d1(p(s(x)))
The TRS R consists of the following rules:
p(s(x)) → x
The set Q consists of the following terms:
d(0)
d(s(x0))
f(s(x0))
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(14) 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].
d(0)
d(s(x0))
f(s(x0))
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
d1(s(x)) → d1(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.
(16) 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 rules of the TRS R:
p(s(x)) → x
Used ordering: Polynomial interpretation [POLO]:
POL(d1(x1)) = x1
POL(p(x1)) = x1
POL(s(x1)) = 2 + x1
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
d1(s(x)) → d1(p(s(x)))
R is empty.
The set Q consists of the following terms:
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(18) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(19) TRUE
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(f, a(s, x)) → A(f, a(p, a(s, x)))
The TRS R consists of the following rules:
a(d, 0) → 0
a(d, a(s, x)) → a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) → a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) → x
The set Q consists of the following terms:
a(d, 0)
a(d, a(s, x0))
a(f, a(s, x0))
a(p, a(s, x0))
We have to consider all minimal (P,Q,R)-chains.
(21) 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.
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(f, a(s, x)) → A(f, a(p, a(s, x)))
The TRS R consists of the following rules:
a(p, a(s, x)) → x
The set Q consists of the following terms:
a(d, 0)
a(d, a(s, x0))
a(f, a(s, x0))
a(p, a(s, x0))
We have to consider all minimal (P,Q,R)-chains.
(23) ATransformationProof (EQUIVALENT transformation)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
f1(s(x)) → f1(p(s(x)))
The TRS R consists of the following rules:
p(s(x)) → x
The set Q consists of the following terms:
d(0)
d(s(x0))
f(s(x0))
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(25) 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].
d(0)
d(s(x0))
f(s(x0))
(26) Obligation:
Q DP problem:
The TRS P consists of the following rules:
f1(s(x)) → f1(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.
(27) 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 rules of the TRS R:
p(s(x)) → x
Used ordering: Polynomial interpretation [POLO]:
POL(f1(x1)) = x1
POL(p(x1)) = x1
POL(s(x1)) = 2 + x1
(28) Obligation:
Q DP problem:
The TRS P consists of the following rules:
f1(s(x)) → f1(p(s(x)))
R is empty.
The set Q consists of the following terms:
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(29) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(30) TRUE