(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(X, g(X)) → f(1, g(X))
g(1) → g(0)
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(X, g(X)) → F(1, g(X))
G(1) → G(0)
The TRS R consists of the following rules:
f(X, g(X)) → f(1, g(X))
g(1) → g(0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(X, g(X)) → F(1, g(X))
The TRS R consists of the following rules:
f(X, g(X)) → f(1, g(X))
g(1) → g(0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(X, g(X)) → F(1, g(X))
The TRS R consists of the following rules:
g(1) → g(0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
F(
X,
g(
X)) →
F(
1,
g(
X)) we obtained the following new rules [LPAR04]:
F(1, g(1)) → F(1, g(1))
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(1, g(1)) → F(1, g(1))
The TRS R consists of the following rules:
g(1) → g(0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
g(1) → g(0)
Used ordering: Knuth-Bendix order [KBO] with precedence:
trivial
and weight map:
F_2=1
g_1=1
0=1
1=2
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(1, g(1)) → F(1, g(1))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) 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(
1,
g(
1)) evaluates to t =
F(
1,
g(
1))
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(1, g(1)) to F(1, g(1)).
(12) NO