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(a, X) → f(X, X)
c → a
c → b
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, X) → f(X, X)
c → a
c → b
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(a, X) → F(X, X)
The TRS R consists of the following rules:
f(a, X) → f(X, X)
c → a
c → b
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ NonTerminationProof
Q DP problem:
The TRS P consists of the following rules:
F(a, X) → F(X, X)
The TRS R consists of the following rules:
f(a, X) → f(X, X)
c → a
c → b
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, X) → F(X, X)
The TRS R consists of the following rules:
f(a, X) → f(X, X)
c → a
c → b
s = F(a, X) evaluates to t =F(X, X)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [X / a]
- Matcher: [ ]
Rewriting sequence
The DP semiunifies directly so there is only one rewrite step from F(a, a) to F(a, a).