NO
Termination w.r.t. Q proof of ../tpdb/TRS/TRCSR/Ex9_Luc06_L.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) → f(b)
b → a
Q is empty.
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
f(X) → f(b)
b → a
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is
b → a
The TRS R 2 is
f(X) → f(b)
The signature Sigma is {f}
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(X) → f(b)
b → a
The set Q consists of the following terms:
f(x0)
b
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) → B
F(X) → F(b)
The TRS R consists of the following rules:
f(X) → f(b)
b → a
The set Q consists of the following terms:
f(x0)
b
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(X) → B
F(X) → F(b)
The TRS R consists of the following rules:
f(X) → f(b)
b → a
The set Q consists of the following terms:
f(x0)
b
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(X) → F(b)
The TRS R consists of the following rules:
f(X) → f(b)
b → a
The set Q consists of the following terms:
f(x0)
b
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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
F(X) → F(b)
The TRS R consists of the following rules:
b → a
The set Q consists of the following terms:
f(x0)
b
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)
↳ 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) → F(b)
The TRS R consists of the following rules:
b → a
The set Q consists of the following terms:
b
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(X) → F(b) at position [0] we obtained the following new rules:
F(X) → F(a)
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(X) → F(a)
The TRS R consists of the following rules:
b → a
The set Q consists of the following terms:
b
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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
F(X) → F(a)
R is empty.
The set Q consists of the following terms:
b
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.
b
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
F(X) → F(a)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(X) → F(a) we obtained the following new rules:
F(a) → F(a)
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Instantiation
↳ QDP
↳ NonTerminationProof
Q DP problem:
The TRS P consists of the following rules:
F(a) → F(a)
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 semiunifying a rule from P directly.
The TRS P consists of the following rules:
F(a) → F(a)
The TRS R consists of the following rules:none
s = F(a) evaluates to t =F(a)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequence
The DP semiunifies directly so there is only one rewrite step from F(a) to F(a).