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) → F(b)
F(X) → 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) → F(b)
F(X) → 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
↳ 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
↳ UsableRulesProof
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
↳ MNOCProof
↳ UsableRulesProof
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.
We use the modular non-overlap check [17] to decrease Q to the empty set.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ NonTerminationProof
↳ UsableRulesProof
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
Q is empty.
We have to consider all (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(X) → F(b)
The TRS R consists of the following rules:
b → a
s = F(X) evaluates to t =F(b)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [X / b]
Rewriting sequence
The DP semiunifies directly so there is only one rewrite step from F(X) to F(b).
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
↳ 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
↳ 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
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
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.