MAYBE

The TRS could not be proven terminating. The proof attempt took 430 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (0ms).
 | – Problem 2 was processed with processor SubtermCriterion (0ms).
 | – Problem 3 was processed with processor SubtermCriterion (0ms).
 | – Problem 4 was processed with processor SubtermCriterion (0ms).
 | – Problem 5 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (52ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (34ms), DependencyGraph (2ms), ReductionPairSAT (25ms), DependencyGraph (1ms), SizeChangePrinciple (8ms)].
 | – Problem 6 was processed with processor SubtermCriterion (0ms).
 | – Problem 7 was processed with processor SubtermCriterion (0ms).
 | – Problem 8 was processed with processor SubtermCriterion (0ms).
 | – Problem 9 was processed with processor SubtermCriterion (0ms).

The following open problems remain:



Open Dependency Pair Problem 5

Dependency Pairs

from#(X)from#(s(X))

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, sel, first, cons, nil


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

sel1#(s(X), cons(Y, Z))sel1#(X, Z)quote#(s(X))quote#(X)
unquote#(s1(X))unquote#(X)first1#(s(X), cons(Y, Z))quote#(Y)
quote#(sel(X, Z))sel1#(X, Z)unquote1#(cons1(X, Z))unquote#(X)
unquote1#(cons1(X, Z))fcons#(unquote(X), unquote1(Z))quote1#(cons(X, Z))quote#(X)
quote1#(first(X, Z))first1#(X, Z)first#(s(X), cons(Y, Z))first#(X, Z)
unquote1#(cons1(X, Z))unquote1#(Z)sel#(s(X), cons(Y, Z))sel#(X, Z)
first1#(s(X), cons(Y, Z))first1#(X, Z)from#(X)from#(s(X))
quote1#(cons(X, Z))quote1#(Z)sel1#(0, cons(X, Z))quote#(X)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons

Strategy


The following SCCs where found

first#(s(X), cons(Y, Z)) → first#(X, Z)

unquote1#(cons1(X, Z)) → unquote1#(Z)

sel#(s(X), cons(Y, Z)) → sel#(X, Z)

first1#(s(X), cons(Y, Z)) → first1#(X, Z)

sel1#(s(X), cons(Y, Z)) → sel1#(X, Z)quote#(s(X)) → quote#(X)
quote#(sel(X, Z)) → sel1#(X, Z)sel1#(0, cons(X, Z)) → quote#(X)

unquote#(s1(X)) → unquote#(X)

from#(X) → from#(s(X))

quote1#(cons(X, Z)) → quote1#(Z)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

unquote#(s1(X))unquote#(X)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

unquote#(s1(X))unquote#(X)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

first1#(s(X), cons(Y, Z))first1#(X, Z)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

first1#(s(X), cons(Y, Z))first1#(X, Z)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

unquote1#(cons1(X, Z))unquote1#(Z)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

unquote1#(cons1(X, Z))unquote1#(Z)

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

quote1#(cons(X, Z))quote1#(Z)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

quote1#(cons(X, Z))quote1#(Z)

Problem 7: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

first#(s(X), cons(Y, Z))first#(X, Z)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

first#(s(X), cons(Y, Z))first#(X, Z)

Problem 8: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

sel#(s(X), cons(Y, Z))sel#(X, Z)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

sel#(s(X), cons(Y, Z))sel#(X, Z)

Problem 9: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

sel1#(s(X), cons(Y, Z))sel1#(X, Z)quote#(s(X))quote#(X)
quote#(sel(X, Z))sel1#(X, Z)sel1#(0, cons(X, Z))quote#(X)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

sel1#(s(X), cons(Y, Z))sel1#(X, Z)quote#(s(X))quote#(X)
quote#(sel(X, Z))sel1#(X, Z)sel1#(0, cons(X, Z))quote#(X)