Termination w.r.t. Q of the following Term Rewriting System could be proven:
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.
↳ QTRS
↳ RRRPoloQTRSProof
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.
The following Q TRS is given: 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.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
a(f, 0) → a(s, 0)
Used ordering:
Polynomial interpretation [25]:
POL(0) = 0
POL(a(x1, x2)) = x1 + x2
POL(d) = 0
POL(f) = 1
POL(p) = 0
POL(s) = 0
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
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.
The TRS is overlay and locally confluent. By [19] we can switch to innermost.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
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))
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
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(d, a(s, x)) → A(s, a(d, a(p, a(s, x))))
A(f, a(s, x)) → A(p, a(s, x))
A(f, a(s, x)) → A(f, a(p, a(s, x)))
A(d, a(s, x)) → A(d, a(p, a(s, x)))
A(d, a(s, x)) → A(s, a(s, 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.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
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(d, a(s, x)) → A(s, a(d, a(p, a(s, x))))
A(f, a(s, x)) → A(p, a(s, x))
A(f, a(s, x)) → A(f, a(p, a(s, x)))
A(d, a(s, x)) → A(d, a(p, a(s, x)))
A(d, a(s, x)) → A(s, a(s, 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.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 5 less nodes.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
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.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
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.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
d(0)
d(s(x0))
f(s(x0))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
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.
By using the rule removal processor [15] with the following polynomial ordering [25], 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: POLO with Polynomial interpretation [25]:
POL(d1(x1)) = x1
POL(p(x1)) = x1
POL(s(x1)) = 1 + x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
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.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
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.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
d(0)
d(s(x0))
f(s(x0))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
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.
By using the rule removal processor [15] with the following polynomial ordering [25], 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: POLO with Polynomial interpretation [25]:
POL(f1(x1)) = x1
POL(p(x1)) = x1
POL(s(x1)) = 1 + x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.