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:
g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
b(b(b(b(x1)))) → a(x1)
Used ordering:
Polynomial interpretation [25]:
POL(a(x1)) = 2 + x1
POL(b(x1)) = 1 + x1
POL(f(x1)) = x1
POL(g(x1)) = 2·x1
POL(h(x1)) = x1
POL(s(x1)) = x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
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(a(a(x1))) → b(a(a(b(x1))))
Used ordering:
Polynomial interpretation [25]:
POL(a(x1)) = 1 + x1
POL(b(x1)) = x1
POL(f(x1)) = x1
POL(g(x1)) = x1
POL(h(x1)) = x1
POL(s(x1)) = x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
b(a(x1)) → a(b(x1))
Used ordering:
Polynomial interpretation [25]:
POL(a(x1)) = 1 + x1
POL(b(x1)) = 2 + 2·x1
POL(f(x1)) = x1
POL(g(x1)) = x1
POL(h(x1)) = x1
POL(s(x1)) = x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F(h(x1)) → H(f(s(h(x1))))
F(h(x1)) → F(s(h(x1)))
F(f(s(s(x1)))) → F(f(x1))
F(s(s(s(x1)))) → H(x1)
F(s(s(s(x1)))) → F(s(h(x1)))
G(h(x1)) → F(s(x1))
G(h(x1)) → G(f(s(x1)))
F(s(s(s(x1)))) → H(f(s(h(x1))))
F(f(s(s(x1)))) → F(x1)
The TRS R consists of the following rules:
g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
F(h(x1)) → H(f(s(h(x1))))
F(h(x1)) → F(s(h(x1)))
F(f(s(s(x1)))) → F(f(x1))
F(s(s(s(x1)))) → H(x1)
F(s(s(s(x1)))) → F(s(h(x1)))
G(h(x1)) → F(s(x1))
G(h(x1)) → G(f(s(x1)))
F(s(s(s(x1)))) → H(f(s(h(x1))))
F(f(s(s(x1)))) → F(x1)
The TRS R consists of the following rules:
g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 5 less nodes.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
F(s(s(s(x1)))) → F(s(h(x1)))
The TRS R consists of the following rules:
g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
F(s(s(s(x1)))) → F(s(h(x1)))
The TRS R consists of the following rules:
h(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ RuleRemovalProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
F(s(s(s(x1)))) → F(s(h(x1)))
The TRS R consists of the following rules:
h(x1) → x1
The set Q consists of the following terms:
h(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 dependency pairs:
F(s(s(s(x1)))) → F(s(h(x1)))
Strictly oriented rules of the TRS R:
h(x1) → x1
Used ordering: POLO with Polynomial interpretation [25]:
POL(F(x1)) = 2·x1
POL(h(x1)) = 1 + x1
POL(s(x1)) = 1 + x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:
h(x0)
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
F(s(s(s(x1)))) → F(s(h(x1)))
The TRS R consists of the following rules:
h(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
F(s(s(s(x1)))) → F(s(h(x1)))
The TRS R consists of the following rules:
h(x1) → x1
The set Q consists of the following terms:
h(x0)
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
F(s(s(s(x1)))) → F(s(h(x1)))
The TRS R consists of the following rules:
h(x1) → x1
Q is empty.
We have to consider all (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
F(f(s(s(x1)))) → F(f(x1))
F(f(s(s(x1)))) → F(x1)
The TRS R consists of the following rules:
g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
F(f(s(s(x1)))) → F(f(x1))
F(f(s(s(x1)))) → F(x1)
The TRS R consists of the following rules:
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
h(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
F(f(s(s(x1)))) → F(f(x1))
F(f(s(s(x1)))) → F(x1)
The TRS R consists of the following rules:
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
h(x1) → x1
Q is empty.
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 dependency pairs:
F(f(s(s(x1)))) → F(x1)
Used ordering: POLO with Polynomial interpretation [25]:
POL(F(x1)) = x1
POL(f(x1)) = 2 + 2·x1
POL(h(x1)) = x1
POL(s(x1)) = x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
F(f(s(s(x1)))) → F(f(x1))
The TRS R consists of the following rules:
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
h(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QDP
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
h(x1) → x1
F(f(s(s(x1)))) → F(f(x1))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
h(x1) → x1
F(f(s(s(x1)))) → F(f(x1))
The set Q is empty.
We have obtained the following QTRS:
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))
The set Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))
The set Q is empty.
We have obtained the following QTRS:
f(s(s(s(x)))) → h(f(s(h(x))))
f(h(x)) → h(f(s(h(x))))
f(f(s(s(x)))) → s(s(s(f(f(x)))))
h(x) → x
F(f(s(s(x)))) → F(f(x))
The set Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(s(s(x)))) → h(f(s(h(x))))
f(h(x)) → h(f(s(h(x))))
f(f(s(s(x)))) → s(s(s(f(f(x)))))
h(x) → x
F(f(s(s(x)))) → F(f(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))
The set Q is empty.
We have obtained the following QTRS:
f(s(s(s(x)))) → h(f(s(h(x))))
f(h(x)) → h(f(s(h(x))))
f(f(s(s(x)))) → s(s(s(f(f(x)))))
h(x) → x
F(f(s(s(x)))) → F(f(x))
The set Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(s(s(x)))) → h(f(s(h(x))))
f(h(x)) → h(f(s(h(x))))
f(f(s(s(x)))) → s(s(s(f(f(x)))))
h(x) → x
F(f(s(s(x)))) → F(f(x))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
H(f(x)) → H(x)
H(f(x)) → H(s(f(h(x))))
S(s(s(f(x)))) → S(f(h(x)))
H(f(x)) → S(f(h(x)))
S(s(f(f(x)))) → S(s(s(x)))
S(s(f(f(x)))) → S(x)
S(s(f(f(x)))) → S(s(x))
S(s(s(f(x)))) → H(x)
S(s(s(f(x)))) → H(s(f(h(x))))
The TRS R consists of the following rules:
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
H(f(x)) → H(x)
H(f(x)) → H(s(f(h(x))))
S(s(s(f(x)))) → S(f(h(x)))
H(f(x)) → S(f(h(x)))
S(s(f(f(x)))) → S(s(s(x)))
S(s(f(f(x)))) → S(x)
S(s(f(f(x)))) → S(s(x))
S(s(s(f(x)))) → H(x)
S(s(s(f(x)))) → H(s(f(h(x))))
The TRS R consists of the following rules:
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))
Q is empty.
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
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
H(f(x)) → H(x)
The TRS R consists of the following rules:
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
H(f(x)) → H(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
H(f(x)) → H(x)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(H(x1)) = 2·x1
POL(f(x1)) = 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
H(f(x)) → H(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
S(s(f(f(x)))) → S(s(s(x)))
S(s(f(f(x)))) → S(x)
S(s(f(f(x)))) → S(s(x))
The TRS R consists of the following rules:
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))
Q is empty.
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 dependency pairs:
S(s(f(f(x)))) → S(s(s(x)))
S(s(f(f(x)))) → S(x)
S(s(f(f(x)))) → S(s(x))
Used ordering: POLO with Polynomial interpretation [25]:
POL(F(x1)) = x1
POL(S(x1)) = 2·x1
POL(f(x1)) = 2 + x1
POL(h(x1)) = x1
POL(s(x1)) = x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
P is empty.
The TRS R consists of the following rules:
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
G(h(x1)) → G(f(s(x1)))
The TRS R consists of the following rules:
g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
G(h(x1)) → G(f(s(x1)))
The TRS R consists of the following rules:
f(s(s(s(x1)))) → h(f(s(h(x1))))
h(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
G(h(x1)) → G(f(s(x1)))
The TRS R consists of the following rules:
f(s(s(s(x1)))) → h(f(s(h(x1))))
h(x1) → x1
Q is empty.
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:
h(x1) → x1
Used ordering: POLO with Polynomial interpretation [25]:
POL(G(x1)) = x1
POL(f(x1)) = x1
POL(h(x1)) = 2 + 2·x1
POL(s(x1)) = 2 + 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ MNOCProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
G(h(x1)) → G(f(s(x1)))
The TRS R consists of the following rules:
f(s(s(s(x1)))) → h(f(s(h(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
G(h(x1)) → G(f(s(x1)))
The TRS R consists of the following rules:
f(s(s(s(x1)))) → h(f(s(h(x1))))
The set Q consists of the following terms:
f(s(s(s(x0))))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule G(h(x1)) → G(f(s(x1))) at position [0] we obtained the following new rules:
G(h(s(s(x0)))) → G(h(f(s(h(x0)))))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
G(h(s(s(x0)))) → G(h(f(s(h(x0)))))
The TRS R consists of the following rules:
f(s(s(s(x1)))) → h(f(s(h(x1))))
The set Q consists of the following terms:
f(s(s(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.
We have reversed the following QTRS:
The set of rules R is
g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)
The set Q is empty.
We have obtained the following QTRS:
h(g(x)) → s(f(g(x)))
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
h(x) → x
s(s(f(f(x)))) → f(f(s(s(s(x)))))
a(b(x)) → b(a(x))
a(a(a(x))) → b(a(a(b(x))))
b(b(b(b(x)))) → a(x)
The set Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
h(g(x)) → s(f(g(x)))
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
h(x) → x
s(s(f(f(x)))) → f(f(s(s(s(x)))))
a(b(x)) → b(a(x))
a(a(a(x))) → b(a(a(b(x))))
b(b(b(b(x)))) → a(x)
Q is empty.