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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 QDP
↳7 QDPOrderProof (⇔)
↳8 QDP
↳9 MNOCProof (⇔)
↳10 QDP
↳11 NonTerminationProof (⇔)
↳12 NO

(0) Obligation:

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

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(app(until, p), f), x) → app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, 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:

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(app(until, p), f), x) → app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(app(until, x0), x1), x2)

(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:

APP(app(app(until, p), f), x) → APP(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x)))
APP(app(app(until, p), f), x) → APP(app(if, app(p, x)), x)
APP(app(app(until, p), f), x) → APP(if, app(p, x))
APP(app(app(until, p), f), x) → APP(p, x)
APP(app(app(until, p), f), x) → APP(app(app(until, p), f), app(f, x))
APP(app(app(until, p), f), x) → APP(f, x)

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(app(until, p), f), x) → app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(app(until, x0), x1), x2)

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 3 less nodes.

(6) Obligation:

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

APP(app(app(until, p), f), x) → APP(app(app(until, p), f), app(f, x))
APP(app(app(until, p), f), x) → APP(p, x)
APP(app(app(until, p), f), x) → APP(f, x)

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(app(until, p), f), x) → app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(app(until, x0), x1), x2)

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(app(until, p), f), x) → APP(p, x)
APP(app(app(until, p), f), x) → APP(f, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x1
app(x1, x2)  =  app(x1, x2)
until  =  until

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

until=1
app_2=1

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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

APP(app(app(until, p), f), x) → APP(app(app(until, p), f), app(f, x))

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(app(until, p), f), x) → app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(app(until, x0), x1), x2)

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

(9) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(10) Obligation:

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

APP(app(app(until, p), f), x) → APP(app(app(until, p), f), app(f, x))

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(app(until, p), f), x) → app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x)))

Q is empty.
We have to consider all (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 = APP(app(app(until, p), f), x) evaluates to t =APP(app(app(until, p), f), app(f, x))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [x / app(f, x)]
  • Semiunifier: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APP(app(app(until, p), f), x) to APP(app(app(until, p), f), app(f, x)).



(12) NO