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

0 QTRS
↳1 AAECC Innermost (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 QDP
↳7 UsableRulesProof (⇔)
↳8 QDP
↳9 QReductionProof (⇔)
↳10 QDP
↳11 Narrowing (⇔)
↳12 QDP
↳13 DependencyGraphProof (⇔)
↳14 QDP
↳15 UsableRulesProof (⇔)
↳16 QDP
↳17 QReductionProof (⇔)
↳18 QDP
↳19 ATransformationProof (⇔)
↳20 QDP
↳21 NonTerminationProof (⇔)
↳22 NO

(0) Obligation:

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

f(tt, x) → f(isList(x), x)
isList(Cons(x, xs)) → isList(xs)
isList(nil) → tt

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

isList(Cons(x, xs)) → isList(xs)
isList(nil) → tt

The TRS R 2 is

f(tt, x) → f(isList(x), x)

The signature Sigma is {f}

(2) Obligation:

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

f(tt, x) → f(isList(x), x)
isList(Cons(x, xs)) → isList(xs)
isList(nil) → tt

The set Q consists of the following terms:

f(tt, x0)
isList(Cons(x0, xs))
isList(nil)

(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(tt, x) → F(isList(x), x)
F(tt, x) → ISLIST(x)
ISLIST(Cons(x, xs)) → ISLIST(xs)

The TRS R consists of the following rules:

f(tt, x) → f(isList(x), x)
isList(Cons(x, xs)) → isList(xs)
isList(nil) → tt

The set Q consists of the following terms:

f(tt, x0)
isList(Cons(x0, xs))
isList(nil)

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 2 less nodes.

(6) Obligation:

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

F(tt, x) → F(isList(x), x)

The TRS R consists of the following rules:

f(tt, x) → f(isList(x), x)
isList(Cons(x, xs)) → isList(xs)
isList(nil) → tt

The set Q consists of the following terms:

f(tt, x0)
isList(Cons(x0, xs))
isList(nil)

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:

F(tt, x) → F(isList(x), x)

The TRS R consists of the following rules:

isList(Cons(x, xs)) → isList(xs)
isList(nil) → tt

The set Q consists of the following terms:

f(tt, x0)
isList(Cons(x0, xs))
isList(nil)

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].

f(tt, x0)

(10) Obligation:

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

F(tt, x) → F(isList(x), x)

The TRS R consists of the following rules:

isList(Cons(x, xs)) → isList(xs)
isList(nil) → tt

The set Q consists of the following terms:

isList(Cons(x0, xs))
isList(nil)

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

(11) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule F(tt, x) → F(isList(x), x) at position [0] we obtained the following new rules [LPAR04]:

F(tt, Cons(x0, xs)) → F(isList(xs), Cons(x0, xs))
F(tt, nil) → F(tt, nil)

(12) Obligation:

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

F(tt, Cons(x0, xs)) → F(isList(xs), Cons(x0, xs))
F(tt, nil) → F(tt, nil)

The TRS R consists of the following rules:

isList(Cons(x, xs)) → isList(xs)
isList(nil) → tt

The set Q consists of the following terms:

isList(Cons(x0, xs))
isList(nil)

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

(13) DependencyGraphProof (EQUIVALENT transformation)

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

(14) Obligation:

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

F(tt, nil) → F(tt, nil)

The TRS R consists of the following rules:

isList(Cons(x, xs)) → isList(xs)
isList(nil) → tt

The set Q consists of the following terms:

isList(Cons(x0, xs))
isList(nil)

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

(15) 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.

(16) Obligation:

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

F(tt, nil) → F(tt, nil)

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

isList(Cons(x0, xs))
isList(nil)

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

(17) 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].

isList(Cons(x0, xs))
isList(nil)

(18) Obligation:

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

F(tt, nil) → F(tt, nil)

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

(19) ATransformationProof (EQUIVALENT transformation)

We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.

(20) Obligation:

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

tt(nil) → tt(nil)

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

(21) 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 = tt(nil) evaluates to t =tt(nil)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from tt(nil) to tt(nil).



(22) NO