MAYBE

The TRS could not be proven terminating. The proof attempt took 407 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 was processed with processor SubtermCriterion (0ms).
 | – 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 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (53ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (35ms), DependencyGraph (1ms), ReductionPairSAT (25ms), DependencyGraph (1ms), SizeChangePrinciple (7ms)].

The following open problems remain:



Open Dependency Pair Problem 9

Dependency Pairs

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

Rewrite Rules

dbl(0)0dbl(s(X))s(s(dbl(X)))
dbls(nil)nildbls(cons(X, Y))cons(dbl(X), dbls(Y))
sel(0, cons(X, Y))Xsel(s(X), cons(Y, Z))sel(X, Z)
indx(nil, X)nilindx(cons(X, Y), Z)cons(sel(X, Z), indx(Y, Z))
from(X)cons(X, from(s(X)))dbl1(0)01
dbl1(s(X))s1(s1(dbl1(X)))sel1(0, cons(X, Y))X
sel1(s(X), cons(Y, Z))sel1(X, Z)quote(0)01
quote(s(X))s1(quote(X))quote(dbl(X))dbl1(X)
quote(sel(X, Y))sel1(X, Y)

Original Signature

Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, nil, cons


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

sel1#(s(X), cons(Y, Z))sel1#(X, Z)quote#(s(X))quote#(X)
dbl1#(s(X))dbl1#(X)sel#(s(X), cons(Y, Z))sel#(X, Z)
dbl#(s(X))dbl#(X)indx#(cons(X, Y), Z)sel#(X, Z)
from#(X)from#(s(X))dbls#(cons(X, Y))dbls#(Y)
dbls#(cons(X, Y))dbl#(X)indx#(cons(X, Y), Z)indx#(Y, Z)
quote#(dbl(X))dbl1#(X)quote#(sel(X, Y))sel1#(X, Y)

Rewrite Rules

dbl(0)0dbl(s(X))s(s(dbl(X)))
dbls(nil)nildbls(cons(X, Y))cons(dbl(X), dbls(Y))
sel(0, cons(X, Y))Xsel(s(X), cons(Y, Z))sel(X, Z)
indx(nil, X)nilindx(cons(X, Y), Z)cons(sel(X, Z), indx(Y, Z))
from(X)cons(X, from(s(X)))dbl1(0)01
dbl1(s(X))s1(s1(dbl1(X)))sel1(0, cons(X, Y))X
sel1(s(X), cons(Y, Z))sel1(X, Z)quote(0)01
quote(s(X))s1(quote(X))quote(dbl(X))dbl1(X)
quote(sel(X, Y))sel1(X, Y)

Original Signature

Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil

Strategy


The following SCCs where found

sel1#(s(X), cons(Y, Z)) → sel1#(X, Z)

quote#(s(X)) → quote#(X)

dbl1#(s(X)) → dbl1#(X)

sel#(s(X), cons(Y, Z)) → sel#(X, Z)

dbl#(s(X)) → dbl#(X)

from#(X) → from#(s(X))

dbls#(cons(X, Y)) → dbls#(Y)

indx#(cons(X, Y), Z) → indx#(Y, Z)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

sel1#(s(X), cons(Y, Z))sel1#(X, Z)

Rewrite Rules

dbl(0)0dbl(s(X))s(s(dbl(X)))
dbls(nil)nildbls(cons(X, Y))cons(dbl(X), dbls(Y))
sel(0, cons(X, Y))Xsel(s(X), cons(Y, Z))sel(X, Z)
indx(nil, X)nilindx(cons(X, Y), Z)cons(sel(X, Z), indx(Y, Z))
from(X)cons(X, from(s(X)))dbl1(0)01
dbl1(s(X))s1(s1(dbl1(X)))sel1(0, cons(X, Y))X
sel1(s(X), cons(Y, Z))sel1(X, Z)quote(0)01
quote(s(X))s1(quote(X))quote(dbl(X))dbl1(X)
quote(sel(X, Y))sel1(X, Y)

Original Signature

Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

sel1#(s(X), cons(Y, Z))sel1#(X, Z)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

dbls#(cons(X, Y))dbls#(Y)

Rewrite Rules

dbl(0)0dbl(s(X))s(s(dbl(X)))
dbls(nil)nildbls(cons(X, Y))cons(dbl(X), dbls(Y))
sel(0, cons(X, Y))Xsel(s(X), cons(Y, Z))sel(X, Z)
indx(nil, X)nilindx(cons(X, Y), Z)cons(sel(X, Z), indx(Y, Z))
from(X)cons(X, from(s(X)))dbl1(0)01
dbl1(s(X))s1(s1(dbl1(X)))sel1(0, cons(X, Y))X
sel1(s(X), cons(Y, Z))sel1(X, Z)quote(0)01
quote(s(X))s1(quote(X))quote(dbl(X))dbl1(X)
quote(sel(X, Y))sel1(X, Y)

Original Signature

Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

dbls#(cons(X, Y))dbls#(Y)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

dbl(0)0dbl(s(X))s(s(dbl(X)))
dbls(nil)nildbls(cons(X, Y))cons(dbl(X), dbls(Y))
sel(0, cons(X, Y))Xsel(s(X), cons(Y, Z))sel(X, Z)
indx(nil, X)nilindx(cons(X, Y), Z)cons(sel(X, Z), indx(Y, Z))
from(X)cons(X, from(s(X)))dbl1(0)01
dbl1(s(X))s1(s1(dbl1(X)))sel1(0, cons(X, Y))X
sel1(s(X), cons(Y, Z))sel1(X, Z)quote(0)01
quote(s(X))s1(quote(X))quote(dbl(X))dbl1(X)
quote(sel(X, Y))sel1(X, Y)

Original Signature

Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

dbl#(s(X))dbl#(X)

Rewrite Rules

dbl(0)0dbl(s(X))s(s(dbl(X)))
dbls(nil)nildbls(cons(X, Y))cons(dbl(X), dbls(Y))
sel(0, cons(X, Y))Xsel(s(X), cons(Y, Z))sel(X, Z)
indx(nil, X)nilindx(cons(X, Y), Z)cons(sel(X, Z), indx(Y, Z))
from(X)cons(X, from(s(X)))dbl1(0)01
dbl1(s(X))s1(s1(dbl1(X)))sel1(0, cons(X, Y))X
sel1(s(X), cons(Y, Z))sel1(X, Z)quote(0)01
quote(s(X))s1(quote(X))quote(dbl(X))dbl1(X)
quote(sel(X, Y))sel1(X, Y)

Original Signature

Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

dbl#(s(X))dbl#(X)

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

indx#(cons(X, Y), Z)indx#(Y, Z)

Rewrite Rules

dbl(0)0dbl(s(X))s(s(dbl(X)))
dbls(nil)nildbls(cons(X, Y))cons(dbl(X), dbls(Y))
sel(0, cons(X, Y))Xsel(s(X), cons(Y, Z))sel(X, Z)
indx(nil, X)nilindx(cons(X, Y), Z)cons(sel(X, Z), indx(Y, Z))
from(X)cons(X, from(s(X)))dbl1(0)01
dbl1(s(X))s1(s1(dbl1(X)))sel1(0, cons(X, Y))X
sel1(s(X), cons(Y, Z))sel1(X, Z)quote(0)01
quote(s(X))s1(quote(X))quote(dbl(X))dbl1(X)
quote(sel(X, Y))sel1(X, Y)

Original Signature

Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

indx#(cons(X, Y), Z)indx#(Y, Z)

Problem 7: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

dbl1#(s(X))dbl1#(X)

Rewrite Rules

dbl(0)0dbl(s(X))s(s(dbl(X)))
dbls(nil)nildbls(cons(X, Y))cons(dbl(X), dbls(Y))
sel(0, cons(X, Y))Xsel(s(X), cons(Y, Z))sel(X, Z)
indx(nil, X)nilindx(cons(X, Y), Z)cons(sel(X, Z), indx(Y, Z))
from(X)cons(X, from(s(X)))dbl1(0)01
dbl1(s(X))s1(s1(dbl1(X)))sel1(0, cons(X, Y))X
sel1(s(X), cons(Y, Z))sel1(X, Z)quote(0)01
quote(s(X))s1(quote(X))quote(dbl(X))dbl1(X)
quote(sel(X, Y))sel1(X, Y)

Original Signature

Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

dbl1#(s(X))dbl1#(X)

Problem 8: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

quote#(s(X))quote#(X)

Rewrite Rules

dbl(0)0dbl(s(X))s(s(dbl(X)))
dbls(nil)nildbls(cons(X, Y))cons(dbl(X), dbls(Y))
sel(0, cons(X, Y))Xsel(s(X), cons(Y, Z))sel(X, Z)
indx(nil, X)nilindx(cons(X, Y), Z)cons(sel(X, Z), indx(Y, Z))
from(X)cons(X, from(s(X)))dbl1(0)01
dbl1(s(X))s1(s1(dbl1(X)))sel1(0, cons(X, Y))X
sel1(s(X), cons(Y, Z))sel1(X, Z)quote(0)01
quote(s(X))s1(quote(X))quote(dbl(X))dbl1(X)
quote(sel(X, Y))sel1(X, Y)

Original Signature

Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

quote#(s(X))quote#(X)