NO
Termination w.r.t. Q proof of ../tpdb/TRS/nontermin/cariboo/ex6.trs
Termination w.r.t. Q of the following Term Rewriting System could be disproven:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, x) → f(i(x), g(g(x)))
f(x, y) → x
g(x) → i(x)
f(x, i(x)) → f(x, x)
f(i(x), i(g(x))) → a
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, x) → f(i(x), g(g(x)))
f(x, y) → x
g(x) → i(x)
f(x, i(x)) → f(x, x)
f(i(x), i(g(x))) → a
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, x) → f(i(x), g(g(x)))
f(x, y) → x
g(x) → i(x)
f(x, i(x)) → f(x, x)
f(i(x), i(g(x))) → a
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
f(x, y) → x
f(i(x), i(g(x))) → a
Used ordering:
Polynomial interpretation [25]:
POL(a) = 1
POL(f(x1, x2)) = 2 + 2·x1 + 2·x2
POL(g(x1)) = x1
POL(i(x1)) = x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, x) → f(i(x), g(g(x)))
g(x) → i(x)
f(x, i(x)) → f(x, x)
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is
g(x) → i(x)
The TRS R 2 is
f(x, x) → f(i(x), g(g(x)))
f(x, i(x)) → f(x, x)
The signature Sigma is {f}
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, x) → f(i(x), g(g(x)))
g(x) → i(x)
f(x, i(x)) → f(x, x)
The set Q consists of the following terms:
f(x0, x0)
g(x0)
f(x0, i(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:
F(x, x) → F(i(x), g(g(x)))
F(x, x) → G(x)
F(x, i(x)) → F(x, x)
F(x, x) → G(g(x))
The TRS R consists of the following rules:
f(x, x) → f(i(x), g(g(x)))
g(x) → i(x)
f(x, i(x)) → f(x, x)
The set Q consists of the following terms:
f(x0, x0)
g(x0)
f(x0, i(x0))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(x, x) → F(i(x), g(g(x)))
F(x, x) → G(x)
F(x, i(x)) → F(x, x)
F(x, x) → G(g(x))
The TRS R consists of the following rules:
f(x, x) → f(i(x), g(g(x)))
g(x) → i(x)
f(x, i(x)) → f(x, x)
The set Q consists of the following terms:
f(x0, x0)
g(x0)
f(x0, i(x0))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(x, x) → F(i(x), g(g(x)))
F(x, i(x)) → F(x, x)
The TRS R consists of the following rules:
f(x, x) → f(i(x), g(g(x)))
g(x) → i(x)
f(x, i(x)) → f(x, x)
The set Q consists of the following terms:
f(x0, x0)
g(x0)
f(x0, i(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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
F(x, x) → F(i(x), g(g(x)))
F(x, i(x)) → F(x, x)
The TRS R consists of the following rules:
g(x) → i(x)
The set Q consists of the following terms:
f(x0, x0)
g(x0)
f(x0, i(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.
f(x0, x0)
f(x0, i(x0))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
F(x, x) → F(i(x), g(g(x)))
F(x, i(x)) → F(x, x)
The TRS R consists of the following rules:
g(x) → i(x)
The set Q consists of the following terms:
g(x0)
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(x, x) → F(i(x), g(g(x))) at position [1] we obtained the following new rules:
F(x, x) → F(i(x), i(g(x)))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
F(x, x) → F(i(x), i(g(x)))
F(x, i(x)) → F(x, x)
The TRS R consists of the following rules:
g(x) → i(x)
The set Q consists of the following terms:
g(x0)
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(x, x) → F(i(x), i(g(x))) at position [1,0] we obtained the following new rules:
F(x, x) → F(i(x), i(i(x)))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(x, x) → F(i(x), i(i(x)))
F(x, i(x)) → F(x, x)
The TRS R consists of the following rules:
g(x) → i(x)
The set Q consists of the following terms:
g(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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
F(x, x) → F(i(x), i(i(x)))
F(x, i(x)) → F(x, x)
R is empty.
The set Q consists of the following terms:
g(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.
g(x0)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
F(x, x) → F(i(x), i(i(x)))
F(x, i(x)) → F(x, x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(x, i(x)) → F(x, x) we obtained the following new rules:
F(i(z0), i(i(z0))) → F(i(z0), i(z0))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
F(x, x) → F(i(x), i(i(x)))
F(i(z0), i(i(z0))) → F(i(z0), i(z0))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(x, x) → F(i(x), i(i(x))) we obtained the following new rules:
F(i(z0), i(z0)) → F(i(i(z0)), i(i(i(z0))))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
F(i(z0), i(z0)) → F(i(i(z0)), i(i(i(z0))))
F(i(z0), i(i(z0))) → F(i(z0), i(z0))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(i(z0), i(i(z0))) → F(i(z0), i(z0)) we obtained the following new rules:
F(i(i(z0)), i(i(i(z0)))) → F(i(i(z0)), i(i(z0)))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
F(i(z0), i(z0)) → F(i(i(z0)), i(i(i(z0))))
F(i(i(z0)), i(i(i(z0)))) → F(i(i(z0)), i(i(z0)))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(i(z0), i(z0)) → F(i(i(z0)), i(i(i(z0)))) we obtained the following new rules:
F(i(i(z0)), i(i(z0))) → F(i(i(i(z0))), i(i(i(i(z0)))))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ NonTerminationProof
Q DP problem:
The TRS P consists of the following rules:
F(i(i(z0)), i(i(z0))) → F(i(i(i(z0))), i(i(i(i(z0)))))
F(i(i(z0)), i(i(i(z0)))) → F(i(i(z0)), i(i(z0)))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
The TRS P consists of the following rules:
F(i(i(z0)), i(i(z0))) → F(i(i(i(z0))), i(i(i(i(z0)))))
F(i(i(z0)), i(i(i(z0)))) → F(i(i(z0)), i(i(z0)))
The TRS R consists of the following rules:none
s = F(i(i(z0')), i(i(i(z0')))) evaluates to t =F(i(i(i(z0'))), i(i(i(i(z0')))))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [z0' / i(z0')]
- Semiunifier: [ ]
Rewriting sequence
F(i(i(z0')), i(i(i(z0')))) → F(i(i(z0')), i(i(z0')))
with rule F(i(i(z0'')), i(i(i(z0'')))) → F(i(i(z0'')), i(i(z0''))) at position [] and matcher [z0'' / z0']
F(i(i(z0')), i(i(z0'))) → F(i(i(i(z0'))), i(i(i(i(z0')))))
with rule F(i(i(z0)), i(i(z0))) → F(i(i(i(z0))), i(i(i(i(z0)))))
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.