Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 UsableRulesProof (⇔)
↳6 QDP
  ↳7 Instantiation (⇔)
    ↳8 QDP
  ↳9 Instantiation (⇔)
    ↳10 QDP
    ↳11 Instantiation (⇔)
    ↳12 QDP
    ↳13 NonTerminationProof (⇔)
    ↳14 FALSE

(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(x) → INF(s(x)) we obtained the following new rules [LPAR04]:

INF(s(z0)) → INF(s(s(z0)))

(10) 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.

(11) 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))))

(12) 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.

(13) 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:
  • Semiunifier: [ ]
  • Matcher: [z0 / s(z0)]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from INF(s(s(z0))) to INF(s(s(s(z0)))).



(14) FALSE