(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a) → f(a)
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
f(a) → f(a)
The signature Sigma is {
f}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a) → f(a)
The set Q consists of the following terms:
f(a)
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a) → F(a)
The TRS R consists of the following rules:
f(a) → f(a)
The set Q consists of the following terms:
f(a)
We have to consider all minimal (P,Q,R)-chains.
(5) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a) → F(a)
R is empty.
The set Q consists of the following terms:
f(a)
We have to consider all minimal (P,Q,R)-chains.
(7) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
f(a)
(8) Obligation:
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.
(9) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
s =
F(
a) evaluates to t =
F(
a)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from F(a) to F(a).
(10) NO