TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60001 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (310ms).
 | – Problem 2 remains open; application of the following processors failed [PolynomialLinearRange4 (283ms), DependencyGraph (1ms), ReductionPairSAT (516ms), DependencyGraph (1ms), SizeChangePrinciple (53335ms)].
 | – Problem 3 remains open; application of the following processors failed [PolynomialLinearRange4 (177ms), DependencyGraph (1ms), ReductionPairSAT (514ms), DependencyGraph (1ms), SizeChangePrinciple (timeout)].
 | – Problem 4 remains open; application of the following processors failed [PolynomialLinearRange4 (279ms), DependencyGraph (51ms), ReductionPairSAT (570ms), DependencyGraph (55ms)].
 | – Problem 5 remains open; application of the following processors failed [PolynomialLinearRange4 (181ms), DependencyGraph (1ms), ReductionPairSAT (440ms), DependencyGraph (1ms)].

The following open problems remain:



Open Dependency Pair Problem 2

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, nil, cons




Open Dependency Pair Problem 3

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, nil, cons




Open Dependency Pair Problem 4

Dependency Pairs

T(sel(x_1, x_2))T(x_2)T(indx(x_1, x_2))T(x_1)
T(s(x_1))T(x_1)sel#(s(X), cons(Y, Z))T(Z)
sel#(s(X), cons(Y, Z))sel#(X, Z)sel#(s(X), cons(Y, Z))T(X)
T(sel(x_1, x_2))T(x_1)T(dbl(x_1))T(x_1)
T(sel(X, Z))sel#(X, Z)T(dbls(x_1))T(x_1)
T(indx(x_1, x_2))T(x_2)sel#(0, cons(X, Y))T(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




Open Dependency Pair Problem 5

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, nil, cons


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

T(indx(x_1, x_2))T(x_1)quote#(s(X))quote#(X)
T(indx(Y, Z))indx#(Y, Z)T(from(s(X)))from#(s(X))
sel1#(s(X), cons(Y, Z))T(X)dbl1#(s(X))T(X)
sel1#(s(X), cons(Y, Z))T(Z)quote#(sel(X, Y))sel1#(X, Y)
dbl1#(s(X))dbl1#(X)quote#(dbl(X))dbl1#(X)
sel#(s(X), cons(Y, Z))sel#(X, Z)sel1#(0, cons(X, Y))T(X)
quote#(s(X))T(X)T(sel(x_1, x_2))T(x_1)
sel1#(s(X), cons(Y, Z))sel1#(X, Z)T(dbl(x_1))T(x_1)
sel#(0, cons(X, Y))T(X)T(sel(x_1, x_2))T(x_2)
T(s(x_1))T(x_1)sel#(s(X), cons(Y, Z))T(Z)
T(dbl(X))dbl#(X)sel#(s(X), cons(Y, Z))T(X)
T(dbls(Y))dbls#(Y)T(dbls(x_1))T(x_1)
T(sel(X, Z))sel#(X, Z)T(indx(x_1, x_2))T(x_2)

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

Context-sensitive strategy:
μ(from#) = μ(from) = μ(01) = μ(T) = μ(0) = μ(s) = μ(nil) = μ(cons) = ∅
μ(quote#) = μ(s1) = μ(dbl1) = μ(dbl) = μ(dbls) = μ(dbls#) = μ(dbl#) = μ(indx) = μ(quote) = μ(dbl1#) = μ(indx#) = {1}
μ(sel#) = μ(sel1#) = μ(sel1) = μ(sel) = {1, 2}


The following SCCs where found

quote#(s(X)) → quote#(X)

sel#(s(X), cons(Y, Z)) → T(X)sel#(s(X), cons(Y, Z)) → sel#(X, Z)
T(sel(x_1, x_2)) → T(x_2)T(indx(x_1, x_2)) → T(x_1)
T(s(x_1)) → T(x_1)T(sel(x_1, x_2)) → T(x_1)
sel#(s(X), cons(Y, Z)) → T(Z)T(dbl(x_1)) → T(x_1)
T(dbls(x_1)) → T(x_1)T(sel(X, Z)) → sel#(X, Z)
sel#(0, cons(X, Y)) → T(X)T(indx(x_1, x_2)) → T(x_2)

sel1#(s(X), cons(Y, Z)) → sel1#(X, Z)

dbl1#(s(X)) → dbl1#(X)