MAYBE

The TRS could not be proven terminating. The proof attempt took 283 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (0ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (53ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (35ms), DependencyGraph (1ms), ReductionPairSAT (21ms), DependencyGraph (2ms), SizeChangePrinciple (4ms)].
 | – 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).

The following open problems remain:



Open Dependency Pair Problem 2

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)))

Original Signature

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


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

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)

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)))

Original Signature

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

Strategy


The following SCCs where found

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 3: 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)))

Original Signature

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

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 4: 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)))

Original Signature

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

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 5: 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)))

Original Signature

Termination of terms over the following signature is verified: 0, s, indx, dbl, from, sel, dbls, 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 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)))

Original Signature

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

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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