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

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 QTRSRRRProof (⇔)
↳4 QTRS
↳5 QTRSRRRProof (⇔)
↳6 QTRS
↳7 Overlay + Local Confluence (⇔)
↳8 QTRS
↳9 DependencyPairsProof (⇔)
↳10 QDP
↳11 UsableRulesProof (⇔)
↳12 QDP
↳13 QReductionProof (⇔)
↳14 QDP
↳15 Instantiation (⇔)
↳16 QDP
↳17 NonTerminationProof (⇔)
↳18 NO

(0) Obligation:

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

app(nil, xs) → nil
app(cons(x, xs), ys) → cons(x, app(xs, ys))
rev(nil) → nil
rev(cons(x, xs)) → append(xs, rev(cons(x, nil)))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(app(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(append(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(nil) = 0   
POL(rev(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

app(nil, xs) → nil


(2) Obligation:

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

app(cons(x, xs), ys) → cons(x, app(xs, ys))
rev(nil) → nil
rev(cons(x, xs)) → append(xs, rev(cons(x, nil)))

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(app(x1, x2)) = x1 + x2   
POL(append(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(nil) = 0   
POL(rev(x1)) = 1 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

rev(nil) → nil


(4) Obligation:

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

app(cons(x, xs), ys) → cons(x, app(xs, ys))
rev(cons(x, xs)) → append(xs, rev(cons(x, nil)))

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(app(x1, x2)) = 2·x1 + 2·x2   
POL(append(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = 2 + x1 + x2   
POL(nil) = 0   
POL(rev(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

app(cons(x, xs), ys) → cons(x, app(xs, ys))


(6) Obligation:

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

rev(cons(x, xs)) → append(xs, rev(cons(x, nil)))

Q is empty.

(7) Overlay + Local Confluence (EQUIVALENT transformation)

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

(8) Obligation:

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

rev(cons(x, xs)) → append(xs, rev(cons(x, nil)))

The set Q consists of the following terms:

rev(cons(x0, x1))

(9) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(10) Obligation:

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

REV(cons(x, xs)) → REV(cons(x, nil))

The TRS R consists of the following rules:

rev(cons(x, xs)) → append(xs, rev(cons(x, nil)))

The set Q consists of the following terms:

rev(cons(x0, x1))

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

(11) 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.

(12) Obligation:

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

REV(cons(x, xs)) → REV(cons(x, nil))

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

rev(cons(x0, x1))

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

(13) 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].

rev(cons(x0, x1))

(14) Obligation:

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

REV(cons(x, xs)) → REV(cons(x, nil))

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

(15) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule REV(cons(x, xs)) → REV(cons(x, nil)) we obtained the following new rules [LPAR04]:

REV(cons(z0, nil)) → REV(cons(z0, nil))

(16) Obligation:

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

REV(cons(z0, nil)) → REV(cons(z0, nil))

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

(17) 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 = REV(cons(z0, nil)) evaluates to t =REV(cons(z0, 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 REV(cons(z0, nil)) to REV(cons(z0, nil)).



(18) NO