(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
cons(x, cons(y, z)) → big
inf(x) → cons(x, inf(s(x)))
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:
INF(x) → CONS(x, inf(s(x)))
INF(x) → INF(s(x))
The TRS R consists of the following rules:
cons(x, cons(y, z)) → big
inf(x) → cons(x, inf(s(x)))
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:
INF(x) → INF(s(x))
The TRS R consists of the following rules:
cons(x, cons(y, z)) → big
inf(x) → cons(x, inf(s(x)))
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:
INF(x) → INF(s(x))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
INF(
x) →
INF(
s(
x)) we obtained the following new rules [LPAR04]:
INF(s(z0)) → INF(s(s(z0)))
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
INF(s(z0)) → INF(s(s(z0)))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
INF(
s(
z0)) →
INF(
s(
s(
z0))) we obtained the following new rules [LPAR04]:
INF(s(s(z0))) → INF(s(s(s(z0))))
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
INF(s(s(z0))) → INF(s(s(s(z0))))
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 =
INF(
s(
s(
z0))) evaluates to t =
INF(
s(
s(
s(
z0))))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [z0 / s(z0)]
- Semiunifier: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from INF(s(s(z0))) to INF(s(s(s(z0)))).
(12) NO