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

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, X1))
from(X) → cons(X, from(s(X)))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, X1))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

2nd(cons1(x0, cons(x1, x2)))
2nd(cons(x0, x1))
from(x0)

(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:

2ND(cons(X, X1)) → 2ND(cons1(X, X1))
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, X1))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

2nd(cons1(x0, cons(x1, x2)))
2nd(cons(x0, x1))
from(x0)

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, X1))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

2nd(cons1(x0, cons(x1, x2)))
2nd(cons(x0, x1))
from(x0)

We have to consider all minimal (P,Q,R)-chains.

(7) 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.

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(X) → FROM(s(X))

R is empty.
The set Q consists of the following terms:

2nd(cons1(x0, cons(x1, x2)))
2nd(cons(x0, x1))
from(x0)

We have to consider all minimal (P,Q,R)-chains.

(9) 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].

2nd(cons1(x0, cons(x1, x2)))
2nd(cons(x0, x1))
from(x0)

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(X) → FROM(s(X))

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 FROM(X) → FROM(s(X)) we obtained the following new rules [LPAR04]:

FROM(s(z0)) → FROM(s(s(z0)))

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(s(z0)) → FROM(s(s(z0)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule FROM(s(z0)) → FROM(s(s(z0))) we obtained the following new rules [LPAR04]:

FROM(s(s(z0))) → FROM(s(s(s(z0))))

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(s(s(z0))) → FROM(s(s(s(z0))))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(15) 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 = FROM(s(s(z0))) evaluates to t =FROM(s(s(s(z0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

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

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Rewriting sequence

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

(16) NO